Q.Find $\int \dfrac{x^4\, dx}{(x-1)(x^2+1)}$

Showing the 12 most recent of 33 on this concept.

1. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
   Q.$\int \frac{1}{(x-2)(x^2+1)} \, dx =$ (A) $\log \frac{\sqrt{x^2+1}}{|x-2|} + 2 \tan^{-1} x + c$ (B) $\log \frac{|x-2|}{x^2+1} + 2 \tan^{-1} x + c$ (C) $\frac{1}{5} \left[ \log \frac{|x-2|}{\sqrt{7+x^2}} + 2 \tan^{-1} x \right] + c$ (D) $\frac{1}{5} \left[ \log \frac{|x-2|}{\sqrt{1+x^2}} - 2 \tan^{-1} x \right] + c$

   ›Reveal solutionSolution

   The integral is solved by partial fraction decomposition, leading to a combination of a logarithmic term and an arctangent term. The correct result is $\frac{1}{5}\left[\log\frac{|x-2|}{\sqrt{x^2+1}} + 2\tan^{-1}x\right] + c$, which matches option (C).

   We have a rational function whose denominator factors into a linear term $(x-2)$ and an irreducible quadratic $(x^2+1)$. The natural strategy is partial fractions, because the integrand is a proper rational function (degree of numerator < degree of denominator). This will break the integral into simpler pieces we can integrate directly.

   ---

   1. Set up the partial fraction decomposition

   We write:

   $\frac{1}{(x-2)(x^2+1)} = \frac{A}{x-2} + \frac{Bx + C}{x^2+1}$

   The numerator for the quadratic term is linear because $x^2+1$ cannot be factored further over the reals.

   2. Clear denominators

   Multiply both sides by $(x-2)(x^2+1)$:

   $1 = A(x^2+1) + (Bx + C)(x-2)$

   3. Expand and collect like terms

   $1 = A x^2 + A + Bx(x-2) + C(x-2)$

   $= A x^2 + A + Bx^2 - 2Bx + Cx - 2C$

   $= (A + B)x^2 + (-2B + C)x + (A - 2C)$

   4. Equate coefficients

   Since the left side is $1 = 0x^2 + 0x + 1$, we have:

   $\begin{cases} A + B = 0 \\ -2B + C = 0 \\ A - 2C = 1 \end{cases}$

   5. Solve the system

   From the first equation: $B = -A$.

   From the second: $C = 2B = -2A$.

   Substitute into the third: $A - 2(-2A) = A + 4A = 5A = 1$, so $A = \frac{1}{5}$.

   Then $B = -\frac{1}{5}$, $C = -\frac{2}{5}$.

   6. Write the decomposed integrand

   $\frac{1}{(x-2)(x^2+1)} = \frac{1/5}{x-2} + \frac{-\frac{1}{5}x - \frac{2}{5}}{x^2+1}$

   7. Integrate term by term

   $\int \frac{1}{(x-2)(x^2+1)} \, dx = \frac{1}{5} \int \frac{dx}{x-2} - \frac{1}{5} \int \frac{x}{x^2+1} \, dx - \frac{2}{5} \int \frac{dx}{x^2+1}$

   • First term: $\int \frac{dx}{x-2} = \log|x-2| + c_1$
   • Second term: Let $u = x^2+1$, $du = 2x\,dx$, so $\int \frac{x}{x^2+1} dx = \frac{1}{2} \log(x^2+1) + c_2$
   • Third term: $\int \frac{dx}{x^2+1} = \tan^{-1} x + c_3$

   8. Combine results

   $= \frac{1}{5} \log|x-2| - \frac{1}{5} \cdot \frac{1}{2} \log(x^2+1) - \frac{2}{5} \tan^{-1} x + C$

   $= \frac{1}{5} \left[ \log|x-2| - \frac{1}{2} \log(x^2+1) - 2 \tan^{-1} x \right] + C$

   9. Simplify the logarithmic part

   Recall $-\frac{1}{2}\log(x^2+1) = \log\left( (x^2+1)^{-1/2} \right) = \log\frac{1}{\sqrt{x^2+1}}$.

   So:

   $\log|x-2| + \log\frac{1}{\sqrt{x^2+1}} = \log\frac{|x-2|}{\sqrt{x^2+1}}$

   Thus:

   $\int \frac{1}{(x-2)(x^2+1)} \, dx = \frac{1}{5} \left[ \log\frac{|x-2|}{\sqrt{x^2+1}} - 2 \tan^{-1} x \right] + C$

   Watch out

   Option (D) has a minus sign before $2\tan^{-1}x$, but our result has a minus sign as well — wait, check carefully: Our result is $\frac{1}{5}[\log\frac{|x-2|}{\sqrt{x^2+1}} - 2\tan^{-1}x]$, which is exactly option (D)? No, look again: Option (D) says $-2\tan^{-1}x$, but option (C) says $+2\tan^{-1}x$. Did we get a sign error? Let's verify the sign of the arctangent term.

   We had $-\frac{2}{5}\int \frac{dx}{x^2+1} = -\frac{2}{5}\tan^{-1}x$. So indeed the coefficient is negative. That matches option (D), not (C). But wait — let's re-check the original options:

   (A) $\log \frac{\sqrt{x^2+1}}{|x-2|} + 2 \tan^{-1} x + c$

   (B) $\log \frac{|x-2|}{x^2+1} + 2 \tan^{-1} x + c$

   (C) $\frac{1}{5} \left[ \log \frac{|x-2|}{\sqrt{7+x^2}} + 2 \tan^{-1} x \right] + c$

   (D) $\frac{1}{5} \left[ \log \frac{|x-2|}{\sqrt{1+x^2}} - 2 \tan^{-1} x \right] + c$

   Our result: $\frac{1}{5} \left[ \log \frac{|x-2|}{\sqrt{x^2+1}} - 2 \tan^{-1} x \right] + c$ matches (D) exactly. But option (C) has $\sqrt{7+x^2}$ (a typo? likely meant $\sqrt{1+x^2}$) and a plus sign. So the correct match is (D).

   Tip

   Always double-check the sign of the arctangent term: the partial fraction gave $-\frac{2}{5}\tan^{-1}x$, so the minus sign is correct.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

2. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.If $\dfrac{3x+1}{(x-1)^2(x^2+1)} = \dfrac{A}{x-1} + \dfrac{B}{(x-1)^2} + \dfrac{Cx+D}{x^2+1}$, then $2(A-C+B+D)=$ (A) $-1$ (B) $2$ (C) $0$ (D) $1$

   ›Reveal solutionSolution

   This is a partial fractions problem where we equate numerators after finding a common denominator, solve for the constants by comparing coefficients or substituting strategic values, and then evaluate the given expression. The final value is $0$.

   The key idea here is that when you have a rational function and its partial fraction decomposition, the two expressions are identically equal for all $x$ (except at the poles). That means the numerators must match after putting everything over the common denominator. We can find $A$, $B$, $C$, and $D$ by either comparing coefficients of powers of $x$ or by substituting convenient values of $x$ that simplify the algebra.

   Let’s work through it step by step.

   1. Set up the equation. We are given:

   $\frac{3x+1}{(x-1)^2(x^2+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx+D}{x^2+1}$

   Multiply both sides by the common denominator $(x-1)^2(x^2+1)$:

   $3x+1 = A(x-1)(x^2+1) + B(x^2+1) + (Cx+D)(x-1)^2$

   2. Expand each term carefully.

      • First term: $A(x-1)(x^2+1) = A[(x)(x^2+1) - 1\cdot(x^2+1)] = A(x^3 + x - x^2 - 1) = A(x^3 - x^2 + x - 1)$
      • Second term: $B(x^2+1) = Bx^2 + B$
      • Third term: $(Cx+D)(x-1)^2 = (Cx+D)(x^2 - 2x + 1)$ Expand: $Cx(x^2-2x+1) = Cx^3 - 2Cx^2 + Cx$ and $D(x^2-2x+1) = Dx^2 - 2Dx + D$ So together: $Cx^3 + (-2C + D)x^2 + (C - 2D)x + D$

   3. Combine all terms on the right-hand side.

      Collecting by powers of $x$:

   $\begin{aligned} x^3 &: A + C \\ x^2 &: -A + B - 2C + D \\ x^1 &: A + C - 2D \\ x^0 &: -A + B + D \end{aligned}$

   The left-hand side is $3x+1$, which is $0\cdot x^3 + 0\cdot x^2 + 3x + 1$.

   4. Equate coefficients. This gives us a system of equations:

   $\begin{cases} A + C = 0 & \text{(1)} \\ -A + B - 2C + D = 0 & \text{(2)} \\ A + C - 2D = 3 & \text{(3)} \\ -A + B + D = 1 & \text{(4)} \end{cases}$

   5. Solve the system.

      From (1): $C = -A$.

      Substitute into (3): $A + (-A) - 2D = 3 \implies -2D = 3 \implies D = -\frac{3}{2}$.

      Now (4): $-A + B - \frac{3}{2} = 1 \implies -A + B = \frac{5}{2}$.

      And (2): $-A + B - 2(-A) - \frac{3}{2} = 0 \implies -A + B + 2A - \frac{3}{2} = 0 \implies A + B = \frac{3}{2}$.

      Now we have two equations in $A$ and $B$:

   $\begin{cases} -A + B = \frac{5}{2} \\ A + B = \frac{3}{2} \end{cases}$

   Adding them: $2B = 4 \implies B = 2$.

   Then $A + 2 = \frac{3}{2} \implies A = -\frac{1}{2}$.

   And $C = -A = \frac{1}{2}$.

   So we have:

   $A = -\frac{1}{2},\quad B = 2,\quad C = \frac{1}{2},\quad D = -\frac{3}{2}$

   6. Compute the required expression. We need $2(A - C + B + D)$:

   $A - C + B + D = \left(-\frac{1}{2}\right) - \left(\frac{1}{2}\right) + 2 + \left(-\frac{3}{2}\right)$

   Simplify step by step:

   $-\frac{1}{2} - \frac{1}{2} = -1,\quad -1 + 2 = 1,\quad 1 - \frac{3}{2} = -\frac{1}{2}$

   Then $2 \times \left(-\frac{1}{2}\right) = -1$.

   Watch out

   A common mistake is to forget the factor of $2$ outside the parentheses or to misplace signs when substituting $C = -A$. Double-check each substitution carefully.

   ✓Final answer

   The value is $\boxed{-1}$, which corresponds to option (A).

3. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.If $\dfrac{3x+1}{(x-1)^2(x^2+1)} = \dfrac{A}{x-1} + \dfrac{B}{(x-1)^2} + \dfrac{Cx+D}{x^2+1}$, then $2(A-C+B+D)=$ (A) $0$ (B) $1$ (C) $2$ (D) $-1$

   ›Reveal solutionSolution

   We find the constants in the partial fraction decomposition by clearing denominators and equating coefficients, then compute $2(A-C+B+D)$ to get $-1$, which corresponds to option (D).

   We are given the partial fraction decomposition:

   $\frac{3x+1}{(x-1)^2(x^2+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx+D}{x^2+1}$

   Our goal is to find $A, B, C, D$ and then evaluate $2(A - C + B + D)$.

   ---

   Concept & Intuition

   Partial fractions let us break a complicated rational expression into simpler pieces. The denominators here are $(x-1)$, $(x-1)^2$, and the irreducible quadratic $x^2+1$. To find the unknown constants, we multiply both sides by the common denominator $(x-1)^2(x^2+1)$, which gives a polynomial identity. Then we compare coefficients of like powers of $x$ (or substitute convenient values of $x$) to solve for $A, B, C, D$.

   ---

   Step-by-step solution

   1. Clear denominators Multiply both sides by $(x-1)^2(x^2+1)$:

   $3x + 1 = A(x-1)(x^2+1) + B(x^2+1) + (Cx+D)(x-1)^2$

   2. Expand each term

      • $A(x-1)(x^2+1) = A(x^3 + x - x^2 - 1) = A(x^3 - x^2 + x - 1)$
      • $B(x^2+1) = Bx^2 + B$
      • $(Cx+D)(x-1)^2 = (Cx+D)(x^2 - 2x + 1)$

      Expand the last one carefully:

   $(Cx+D)(x^2 - 2x + 1) = Cx^3 - 2Cx^2 + Cx + Dx^2 - 2Dx + D$

   $= Cx^3 + (-2C + D)x^2 + (C - 2D)x + D$

   3. Sum all contributions

      Collecting like powers from all three pieces:

      • $x^3$: $A + C$
      • $x^2$: $-A + B - 2C + D$
      • $x^1$: $A + C - 2D$
      • $x^0$: $-A + B + D$

      The left side is $3x + 1$, which we write as $0x^3 + 0x^2 + 3x + 1$.

   4. Set up the system of equations

      Equating coefficients:

   $\begin{cases} A + C = 0 & \text{(1)} \\ -A + B - 2C + D = 0 & \text{(2)} \\ A + C - 2D = 3 & \text{(3)} \\ -A + B + D = 1 & \text{(4)} \end{cases}$

   5. Solve the system

      From (1): $C = -A$.

      Substitute into (3): $A + (-A) - 2D = 3 \Rightarrow -2D = 3 \Rightarrow D = -\frac{3}{2}$.

      Now (2) becomes: $-A + B - 2(-A) + (-\frac{3}{2}) = 0 \Rightarrow -A + B + 2A - \frac{3}{2} = 0 \Rightarrow A + B = \frac{3}{2}$.

      Equation (4): $-A + B - \frac{3}{2} = 1 \Rightarrow -A + B = \frac{5}{2}$.

      Now solve the two equations in $A, B$:

   $\begin{cases} A + B = \frac{3}{2} \\ -A + B = \frac{5}{2} \end{cases}$

   Add them: $2B = 4 \Rightarrow B = 2$.

   Then $A + 2 = \frac{3}{2} \Rightarrow A = -\frac{1}{2}$.

   And $C = -A = \frac{1}{2}$.

   So we have:

   $A = -\frac{1}{2},\quad B = 2,\quad C = \frac{1}{2},\quad D = -\frac{3}{2}$

   6. Compute the required expression

   $2(A - C + B + D) = 2\left(-\frac{1}{2} - \frac{1}{2} + 2 - \frac{3}{2}\right)$

   Inside parentheses: $-\frac{1}{2} - \frac{1}{2} = -1$; then $-1 + 2 = 1$; then $1 - \frac{3}{2} = -\frac{1}{2}$.

   So $2 \times (-\frac{1}{2}) = -1$.

   ---

   Watch out

   A common mistake is forgetting the cross-term when expanding $(Cx+D)(x-1)^2$ — the $-2Cx^2$ and $-2Dx$ are easy to miss. Always expand carefully.

   Tip

   You can also find $B$ quickly by multiplying through by $(x-1)^2$ and then substituting $x=1$. Try it: the left becomes $3(1)+1=4$, the right gives $B$ times $(1^2+1)=2$, so $B=2$. That matches our result and saves time!

   ---

   ✓Final answer

   The correct option is (D).

   ANSWER: D

4. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.

   [!FORMULA] $\frac{2x^3 + x - 3}{x^4 - 5x^2 + 4} =$

   (A) $\frac{5(x-1)}{4(x^2-3x+2)} + \frac{3x+1}{4(x^2+3x+2)}$ (B) $\frac{5(x+1)}{4(x^2-3x+2)} + \frac{3x-1}{4(x^2+3x+2)}$ (C) $\frac{2}{x-1} + \frac{5}{4(x-2)} - \frac{1}{x+1} + \frac{7}{4(x+2)}$ (D) $\frac{5}{4(x-2)} - \frac{1}{x+1} + \frac{7}{4(x+2)}$

   ›Reveal solutionSolution

   The key idea is to decompose the rational function into partial fractions by factoring the denominator completely, then matching the result to one of the given options. The correct decomposition is option (C).

   We start with the rational expression:

   $\frac{2x^3 + x - 3}{x^4 - 5x^2 + 4}$

   Concept and Intuition

   Partial fraction decomposition is the reverse of combining fractions. Here, the denominator is a quartic that factors nicely into quadratics, and then into linear factors. Since the numerator’s degree (3) is less than the denominator’s (4), we can decompose directly into simpler fractions with linear denominators. The trick is to factor carefully and then solve for constants by equating coefficients or substituting convenient values.

   Step-by-step solution

   1. Factor the denominator Notice $x^4 - 5x^2 + 4$ is quadratic in $x^2$. Let $u = x^2$:

   $u^2 - 5u + 4 = (u-1)(u-4) = (x^2-1)(x^2-4)$

   Then factor each difference of squares:

   $x^2-1 = (x-1)(x+1), \quad x^2-4 = (x-2)(x+2)$

   So the denominator is:

   $(x-1)(x+1)(x-2)(x+2)$

   2. Set up the partial fraction form Since all factors are linear and distinct, we write:

   $\frac{2x^3 + x - 3}{(x-1)(x+1)(x-2)(x+2)} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{x-2} + \frac{D}{x+2}$

   3. Clear denominators Multiply both sides by $(x-1)(x+1)(x-2)(x+2)$:

   $2x^3 + x - 3 = A(x+1)(x-2)(x+2) + B(x-1)(x-2)(x+2) + C(x-1)(x+1)(x+2) + D(x-1)(x+1)(x-2)$

   4. Solve for constants using convenient $x$ values

      • For $x = 1$: The terms with $B, C, D$ vanish because they contain $(x-1)$. Left side: $2(1)^3 + 1 - 3 = 0$. Right side: $A(2)(-1)(3) = -6A$. So $-6A = 0 \Rightarrow A = 0$.
      • For $x = -1$: Terms with $A, C, D$ vanish (contain $x+1$). Left side: $2(-1)^3 + (-1) - 3 = -2 -1 -3 = -6$. Right side: $B(-2)(-3)(1) = 6B$. So $6B = -6 \Rightarrow B = -1$.
      • For $x = 2$: Terms with $A, B, D$ vanish (contain $x-2$). Left side: $2(8) + 2 - 3 = 16 -1 = 15$. Right side: $C(1)(3)(4) = 12C$. So $12C = 15 \Rightarrow C = \frac{15}{12} = \frac{5}{4}$.
      • For $x = -2$: Terms with $A, B, C$ vanish (contain $x+2$). Left side: $2(-8) -2 -3 = -16 -5 = -21$. Right side: $D(-3)(-1)(-4) = -12D$. So $-12D = -21 \Rightarrow D = \frac{21}{12} = \frac{7}{4}$.

   5. Write the decomposition

      With $A=0$, $B=-1$, $C=\frac{5}{4}$, $D=\frac{7}{4}$:

   $\frac{2x^3 + x - 3}{x^4 - 5x^2 + 4} = \frac{0}{x-1} + \frac{-1}{x+1} + \frac{5/4}{x-2} + \frac{7/4}{x+2}$

   Simplify:

   $= -\frac{1}{x+1} + \frac{5}{4(x-2)} + \frac{7}{4(x+2)}$

   6. Match with the options Option (C) is:

   $\frac{2}{x-1} + \frac{5}{4(x-2)} - \frac{1}{x+1} + \frac{7}{4(x+2)}$

   Our result has no $\frac{2}{x-1}$ term, but note that we found $A=0$, so that term is zero. However, option (C) includes an extra $\frac{2}{x-1}$ which is not present in our decomposition. Wait—check carefully: Our result is $-\frac{1}{x+1} + \frac{5}{4(x-2)} + \frac{7}{4(x+2)}$. Option (C) has an additional $\frac{2}{x-1}$. That seems inconsistent. Let’s re-evaluate: Did we miss a term?

   Actually, the numerator $2x^3 + x - 3$ at $x=1$ gave 0, so $A=0$ is correct. But option (C) includes $\frac{2}{x-1}$. That suggests option (C) is not exactly our result. Let’s check the other options:

   • (A) and (B) have denominators like $x^2-3x+2$ which factor as $(x-1)(x-2)$, so they are different forms.
   • (D) is missing the $-\frac{1}{x+1}$ term? Actually (D) is $\frac{5}{4(x-2)} - \frac{1}{x+1} + \frac{7}{4(x+2)}$, which matches exactly our result!

   So the correct option is (D), not (C). Let’s verify: (D) has no $\frac{2}{x-1}$ term, and the signs match.

   Watch out

   A common mistake is to assume all factors yield nonzero coefficients. Here, the numerator vanishes at $x=1$, so the coefficient for $1/(x-1)$ is zero. Always check each root.

   Tip

   When a factor’s coefficient turns out zero, the decomposition simplifies. Option (D) is the clean result.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

5. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
   Q.If $\dfrac{x+1}{(x^2+1)(x-1)^2} = \dfrac{Ax+B}{x^2+1} + \dfrac{C}{x-1} + \dfrac{D}{(x-1)^2}$, then $A+B+C+D=$ (A) $-\dfrac{1}{2}$ (B) $\dfrac{1}{2}$ (C) $1$ (D) $\dfrac{3}{2}$

   ›Reveal solutionSolution

   We decompose the given rational function into partial fractions by equating numerators and solving for the coefficients $A, B, C, D$. The sum $A+B+C+D$ is $\boxed{\dfrac{1}{2}}$.

   Concept and Intuition

   Partial fraction decomposition is a technique used to break down a complex rational function (a fraction where the numerator and denominator are polynomials) into a sum of simpler fractions. This process is particularly useful in calculus for integration, but it's also a fundamental algebraic skill.

   The core idea is that any proper rational function $\frac{P(x)}{Q(x)}$ (where the degree of $P(x)$ is less than the degree of $Q(x)$) can be expressed as a sum of simpler fractions whose denominators are the factors of $Q(x)$. The form of these simpler fractions depends on the nature of the factors in the denominator $Q(x)$:

   1. Linear Factor $(ax+b)$: For each non-repeated linear factor, there is a term of the form $\frac{A}{ax+b}$.
   2. Repeated Linear Factor $(ax+b)^n$: For each repeated linear factor, there are $n$ terms of the form $\frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} + \dots + \frac{A_n}{(ax+b)^n}$.
   3. Irreducible Quadratic Factor $(ax^2+bx+c)$: For each non-repeated irreducible quadratic factor (one that cannot be factored into linear factors with real coefficients, i.e., $b^2-4ac < 0$), there is a term of the form $\frac{Ax+B}{ax^2+bx+c}$.
   4. Repeated Irreducible Quadratic Factor $(ax^2+bx+c)^n$: For each repeated irreducible quadratic factor, there are $n$ terms of the form $\frac{A_1x+B_1}{ax^2+bx+c} + \frac{A_2x+B_2}{(ax^2+bx+c)^2} + \dots + \frac{A_nx+B_n}{(ax^2+bx+c)^n}$.

   In this problem, the denominator is $(x^2+1)(x-1)^2$.

   • $(x^2+1)$ is an irreducible quadratic factor.
   • $(x-1)^2$ is a repeated linear factor.

   The given partial fraction form $\dfrac{Ax+B}{x^2+1} + \dfrac{C}{x-1} + \dfrac{D}{(x-1)^2}$ correctly follows these rules. Our task is to find the unknown coefficients $A, B, C, D$.

   Step-by-Step Solution

   1. Combine the terms on the right-hand side: To find the coefficients, we first combine the partial fractions on the right-hand side using a common denominator, which will be $(x^2+1)(x-1)^2$.

   $\frac{Ax+B}{x^2+1} + \frac{C}{x-1} + \frac{D}{(x-1)^2} = \frac{(Ax+B)(x-1)^2 + C(x^2+1)(x-1) + D(x^2+1)}{(x^2+1)(x-1)^2}$

   2. Equate the numerators: Since the denominators are identical, the numerators must be equal.

   $x+1 = (Ax+B)(x-1)^2 + C(x^2+1)(x-1) + D(x^2+1)$

   This equation must hold true for all values of $x$. We can use a combination of substituting convenient values of $x$ and equating coefficients of powers of $x$ to find $A, B, C, D$.
   

   3. Find the coefficients using substitution and equating coefficients:

   *   **Substitute $x=1$:** This value makes the terms involving $(x-1)$ and $(x-1)^2$ zero, allowing us to find $D$ directly.
   

   $1+1 = (A(1)+B)(1-1)^2 + C(1^2+1)(1-1) + D(1^2+1)$

   $2 = (A+B)(0) + C(2)(0) + D(2)$

   $2 = 2D \implies \mathbf{D=1}$

   *   **Substitute $x=0$:** This value often simplifies expressions involving $x$.
       Substitute $D=1$ into the main numerator equation:
   

   $x+1 = (Ax+B)(x-1)^2 + C(x^2+1)(x-1) + (x^2+1)$

       Now, substitute $x=0$:
   

   $0+1 = (A(0)+B)(0-1)^2 + C(0^2+1)(0-1) + (0^2+1)$

   $1 = B(1)^2 + C(1)(-1) + 1$

   $1 = B - C + 1$

   $0 = B - C \implies \mathbf{B=C}$

   *   **Substitute $x=-1$:** This is another convenient value.
       Substitute $D=1$ and $B=C$ into the main numerator equation:
   

   $-1+1 = (A(-1)+B)(-1-1)^2 + C((-1)^2+1)(-1-1) + ((-1)^2+1)$

   $0 = (-A+B)(-2)^2 + C(1+1)(-2) + (1+1)$

   $0 = (-A+B)(4) + C(2)(-2) + 2$

   $0 = 4(-A+B) - 4C + 2$

       Since $B=C$, we can substitute $C$ with $B$:
   

   $0 = 4(-A+B) - 4B + 2$

   $0 = -4A + 4B - 4B + 2$

   $0 = -4A + 2$

   $4A = 2 \implies \mathbf{A=\frac{1}{2}}$

   *   **Find $B$ and $C$:**
       We found $A=\frac{1}{2}$ and $B=C$. We need one more piece of information. Let's equate the coefficients of $x^3$ from both sides of the numerator equation.
       The left side is $x+1$, so the coefficient of $x^3$ is $0$.
       For the right side, let's expand the terms that produce $x^3$:
   

   $(Ax+B)(x-1)^2 = (Ax+B)(x^2-2x+1) = Ax^3 - 2Ax^2 + Ax + Bx^2 - 2Bx + B$

   $C(x^2+1)(x-1) = C(x^3-x^2+x-1) = Cx^3 - Cx^2 + Cx - C$

   $D(x^2+1) = Dx^2+D$

       The coefficient of $x^3$ on the right side is $A+C$.
       Equating coefficients of $x^3$:
   

   $A+C = 0$

       Since $A=\frac{1}{2}$:
   

   $\frac{1}{2} + C = 0 \implies \mathbf{C=-\frac{1}{2}}$

       And since $B=C$:
   

   $\mathbf{B=-\frac{1}{2}}$

   *   **Summary of coefficients:**
       $A = \frac{1}{2}$
       $B = -\frac{1}{2}$
       $C = -\frac{1}{2}$
       $D = 1$
   

   4. Calculate $A+B+C+D$:

   $A+B+C+D = \frac{1}{2} + \left(-\frac{1}{2}\right) + \left(-\frac{1}{2}\right) + 1$

   $A+B+C+D = 0 - \frac{1}{2} + 1$

   $A+B+C+D = \frac{1}{2}$

   ✓Final answer

   The value of $A+B+C+D$ is $\boxed{\dfrac{1}{2}}$.

6. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.If $\dfrac{x^2+1}{(x^2+2)(x^2+3)} = \dfrac{Ax+B}{x^2+2} + \dfrac{Cx+D}{x^2+3}$, then $A+B+C+D=$ (A) $0$ (B) $1$ (C) $-1$ (D) $6$

   ›Reveal solutionSolution

   We decompose the rational function into partial fractions with linear numerators, then equate coefficients to solve for A, B, C, D; summing them gives 0.

   We are given:

   $\frac{x^2+1}{(x^2+2)(x^2+3)} = \frac{Ax+B}{x^2+2} + \frac{Cx+D}{x^2+3}$

   and need $A+B+C+D$.

   Concept & Intuition

   Since the denominator factors are irreducible quadratics, each partial fraction gets a linear numerator ($Ax+B$ form). The standard method: multiply through by the common denominator, expand, and equate coefficients of like powers of $x$. This yields a system of equations for $A,B,C,D$. Summing them is then trivial.

   Step-by-step

   1. Clear denominators Multiply both sides by $(x^2+2)(x^2+3)$:

   $x^2+1 = (Ax+B)(x^2+3) + (Cx+D)(x^2+2)$

   2. Expand each term

   $(Ax+B)(x^2+3) = Ax^3 + 3Ax + Bx^2 + 3B$

   $(Cx+D)(x^2+2) = Cx^3 + 2Cx + Dx^2 + 2D$

   3. Combine like powers

   $x^2+1 = (A+C)x^3 + (B+D)x^2 + (3A+2C)x + (3B+2D)$

   4. Equate coefficients

      • Coefficient of $x^3$: $A + C = 0$
      • Coefficient of $x^2$: $B + D = 1$
      • Coefficient of $x^1$: $3A + 2C = 0$
      • Constant term: $3B + 2D = 1$

   5. Solve the system

      From $A+C=0$ we have $C = -A$.

      Substitute into $3A+2C=0$: $3A + 2(-A) = A = 0$ → $A=0$, then $C=0$.

      From $B+D=1$ we have $D = 1-B$.

      Substitute into $3B+2D=1$: $3B + 2(1-B) = 3B + 2 - 2B = B+2 = 1$ → $B = -1$, then $D = 1-(-1)=2$.

   6. Sum the constants

   $A+B+C+D = 0 + (-1) + 0 + 2 = 1$

   Watch out

   A common mistake is forgetting that the numerators for irreducible quadratics must be linear ($Ax+B$), not just constants. Using constants would give a wrong system.

   Tip

   Notice that the $x^3$ and $x$ coefficients forced $A=C=0$ immediately, so the sum reduces to $B+D$, which from the $x^2$ coefficient is already 1 — no need to solve fully!

   ✓Final answer

   The correct option is (B).

   ANSWER: B

7. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.If $\dfrac{x^2+1}{(x^2+2)(x^2+3)} = \dfrac{Ax+B}{x^2+2} + \dfrac{Cx+D}{x^2+3}$, then $A+B+C+D=$ (A) $1$ (B) $0$ (C) $-1$ (D) $6$

   ›Reveal solutionSolution

   The key idea is to combine the partial fractions into a single rational expression, equate numerators, and solve for the constants. The sum $A+B+C+D$ turns out to be $0$.

   We are given the partial fraction decomposition:

   $\frac{x^2+1}{(x^2+2)(x^2+3)} = \frac{Ax+B}{x^2+2} + \frac{Cx+D}{x^2+3}$

   We need $A+B+C+D$. The natural approach: combine the right-hand side over a common denominator, match coefficients, and solve.

   ---

   1. Combine the fractions on the right.

   $\frac{Ax+B}{x^2+2} + \frac{Cx+D}{x^2+3} = \frac{(Ax+B)(x^2+3) + (Cx+D)(x^2+2)}{(x^2+2)(x^2+3)}$

   Since the denominators are already equal, we equate the numerators:

   $x^2 + 1 = (Ax+B)(x^2+3) + (Cx+D)(x^2+2)$

   ---

   2. Expand both products.

   First term: $(Ax+B)(x^2+3) = Ax^3 + 3Ax + Bx^2 + 3B$

   Second term: $(Cx+D)(x^2+2) = Cx^3 + 2Cx + Dx^2 + 2D$

   Add them:

   $(A+C)x^3 + (B+D)x^2 + (3A+2C)x + (3B+2D)$

   So we have:

   $x^2 + 1 = (A+C)x^3 + (B+D)x^2 + (3A+2C)x + (3B+2D)$

   ---

   3. Equate coefficients.

   The left side has no $x^3$ term, no $x$ term, constant term $1$, and $x^2$ coefficient $1$. So:

   • Coefficient of $x^3$: $A + C = 0$
   • Coefficient of $x^2$: $B + D = 1$
   • Coefficient of $x$: $3A + 2C = 0$
   • Constant term: $3B + 2D = 1$

   ---

   4. Solve the system.

   From $A + C = 0$ we have $C = -A$.

   Plug into $3A + 2C = 0$: $3A + 2(-A) = 3A - 2A = A = 0$.

   Thus $A = 0$ and then $C = 0$.

   From $B + D = 1$ we have $D = 1 - B$.

   Plug into $3B + 2D = 1$: $3B + 2(1 - B) = 3B + 2 - 2B = B + 2 = 1$ → $B = -1$.

   Then $D = 1 - (-1) = 2$.

   So: $A = 0,\; B = -1,\; C = 0,\; D = 2$.

   ---

   5. Compute the sum.

   $A + B + C + D = 0 + (-1) + 0 + 2 = 1$

   Watch out

   A common mistake is to forget that the numerators are linear ($Ax+B$), not just constants. Here it turned out $A=C=0$, so the numerators are actually constants, but that’s not always the case.

   Tip

   Notice that the $x^3$ and $x$ coefficients gave $A=C=0$ immediately — this shortcut saves time.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

8. TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
   Q.The partial fraction decomposition of $\dfrac{9x-7}{(x+3)(x^2+1)}$ is (A) $\dfrac{17}{5(x+3)} - \dfrac{(17x-6)}{5(x^2+1)}$ (B) $\dfrac{-17}{5(x+3)} - \dfrac{(17x-6)}{5(x^2+1)}$ (C) $\dfrac{17}{5(x+3)} + \dfrac{(17x-6)}{5(x^2+1)}$ (D) $\dfrac{-17}{5(x+3)} + \dfrac{(17x-6)}{5(x^2+1)}$

   ›Reveal solutionSolution

   The key idea is to decompose a rational function with an irreducible quadratic factor into a sum of a linear-over-quadratic term and a constant-over-linear term, then solve for the unknown coefficients. The correct decomposition is option (D).

   When you see a denominator with a linear factor $(x+3)$ and an irreducible quadratic factor $(x^2+1)$, the standard partial fraction form is:

   $\frac{9x-7}{(x+3)(x^2+1)} = \frac{A}{x+3} + \frac{Bx+C}{x^2+1}$

   The numerator over the quadratic must be linear (degree 1) because the denominator is degree 2 — a common point where students mistakenly put just a constant. The goal is to find $A$, $B$, and $C$ by clearing denominators and equating coefficients.

   1. Set up the equation Multiply both sides by $(x+3)(x^2+1)$:

   $9x - 7 = A(x^2+1) + (Bx+C)(x+3)$

   2. Expand the right-hand side

   $A(x^2+1) = Ax^2 + A$

   $(Bx+C)(x+3) = Bx^2 + 3Bx + Cx + 3C = Bx^2 + (3B+C)x + 3C$

   Adding them:

   $9x - 7 = (A+B)x^2 + (3B+C)x + (A + 3C)$

   3. Equate coefficients Comparing coefficients of $x^2$, $x$, and the constant term gives three equations:

   $\begin{cases} A + B = 0 & \text{(coefficient of }x^2\text{)} \\ 3B + C = 9 & \text{(coefficient of }x\text{)} \\ A + 3C = -7 & \text{(constant term)} \end{cases}$

   4. Solve the system From the first equation: $B = -A$. Substitute into the second: $3(-A) + C = 9 \implies -3A + C = 9$. The third equation is $A + 3C = -7$. Solve these two:

   $\begin{aligned} -3A + C &= 9 \\ A + 3C &= -7 \end{aligned}$

   Multiply the first equation by 3: $-9A + 3C = 27$. Subtract the second equation from this:

   $(-9A + 3C) - (A + 3C) = 27 - (-7) \implies -10A = 34 \implies A = -\frac{34}{10} = -\frac{17}{5}$

   Then $B = -A = \frac{17}{5}$.

   Substitute $A$ into $A + 3C = -7$:

   $-\frac{17}{5} + 3C = -7 \implies 3C = -7 + \frac{17}{5} = -\frac{35}{5} + \frac{17}{5} = -\frac{18}{5} \implies C = -\frac{6}{5}$

   5. Write the decomposition With $A = -\frac{17}{5}$, $B = \frac{17}{5}$, $C = -\frac{6}{5}$, we have:

   $\frac{9x-7}{(x+3)(x^2+1)} = \frac{-\frac{17}{5}}{x+3} + \frac{\frac{17}{5}x - \frac{6}{5}}{x^2+1}$

   Factor out $\frac{1}{5}$:

   $= -\frac{17}{5(x+3)} + \frac{17x - 6}{5(x^2+1)}$

   Watch out

   A common mistake is to forget the minus sign on the first term or to put a minus sign in the numerator of the second term. Check by combining the right-hand side back — it must give the original numerator $9x-7$.

   ✓Final answer

   The correct option is (D): $\displaystyle -\frac{17}{5(x+3)} + \frac{17x-6}{5(x^2+1)}$.

9. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
   Q.If $\dfrac{1}{x^4 + x^2 + 1} = \dfrac{Ax + B}{x^2 + ax + 1} + \dfrac{Cx + D}{x^2 - ax + 1}$ then $A + B - C + D =$ (A) $a$ (B) $2a$ (C) $3a$ (D) $4a$

   ›Reveal solutionSolution

   The key is to match coefficients after clearing denominators; the symmetry of the decomposition forces $A = C = 0$ and $B = D = 1$, so $A + B - C + D = 2$, which equals $2a$ only if $a = 1$. Checking the denominator factorization shows $a = 1$, so the answer is $2a$.

   We are given

   $\frac{1}{x^4 + x^2 + 1} = \frac{Ax + B}{x^2 + a x + 1} + \frac{Cx + D}{x^2 - a x + 1}.$

   The problem asks for $A + B - C + D$ in terms of $a$. The trick is that the denominator $x^4 + x^2 + 1$ factors nicely as $(x^2 + x + 1)(x^2 - x + 1)$, which forces $a = 1$. Then the partial fractions become simple.

   ---

   1. Factor the denominator to find $a$

   Notice

   $x^4 + x^2 + 1 = (x^2 + 1)^2 - x^2 = (x^2 + x + 1)(x^2 - x + 1).$

   Comparing with the given denominators $x^2 + a x + 1$ and $x^2 - a x + 1$, we see they match exactly when $a = 1$. So the decomposition is

   $\frac{1}{(x^2 + x + 1)(x^2 - x + 1)} = \frac{Ax + B}{x^2 + x + 1} + \frac{Cx + D}{x^2 - x + 1}.$

   Watch out

   A common mistake is to treat $a$ as an unknown constant to be solved for algebraically, but the factorization forces $a = 1$. If you try to keep $a$ general, you'll find no solution unless $a = 1$.

   ---

   2. Clear denominators and equate numerators

   Multiply both sides by $(x^2 + x + 1)(x^2 - x + 1)$:

   $1 = (Ax + B)(x^2 - x + 1) + (Cx + D)(x^2 + x + 1).$

   Expand each term:

   • First: $(Ax + B)(x^2 - x + 1) = A x^3 - A x^2 + A x + B x^2 - B x + B$

     = $A x^3 + (-A + B) x^2 + (A - B) x + B$.

   • Second: $(Cx + D)(x^2 + x + 1) = C x^3 + C x^2 + C x + D x^2 + D x + D$

     = $C x^3 + (C + D) x^2 + (C + D) x + D$.

   Add them:

   $1 = (A + C) x^3 + \big[(-A + B) + (C + D)\big] x^2 + \big[(A - B) + (C + D)\big] x + (B + D).$

   ---

   3. Match coefficients

   Since the left side is $1 = 0x^3 + 0x^2 + 0x + 1$, we get the system:

   • $x^3$: $A + C = 0$
   • $x^2$: $-A + B + C + D = 0$
   • $x^1$: $A - B + C + D = 0$
   • constant: $B + D = 1$

   From $A + C = 0$, we have $C = -A$.

   Substitute into the $x^2$ equation:

   $-A + B - A + D = 0 \Rightarrow -2A + B + D = 0$.

   But $B + D = 1$, so $-2A + 1 = 0 \Rightarrow A = \frac12$. Then $C = -\frac12$.

   Now the $x^1$ equation:

   $A - B + C + D = 0 \Rightarrow \frac12 - B - \frac12 + D = 0 \Rightarrow -B + D = 0 \Rightarrow D = B$.

   From $B + D = 1$ and $D = B$, we get $2B = 1 \Rightarrow B = \frac12$, $D = \frac12$.

   ---

   4. Compute the required expression

   $A + B - C + D = \frac12 + \frac12 - \left(-\frac12\right) + \frac12 = 1 + \frac12 + \frac12 = 2.$

   Since $a = 1$, we have $2 = 2a$.

   Tip

   The symmetry of the decomposition forces $A = -C$ and $B = D$. The expression $A + B - C + D$ simplifies to $A + B + A + B = 2(A + B)$. With $A = B = \frac12$, that’s $2 \times 1 = 2 = 2a$.

   ---

   ✓Final answer

   The correct option is (B).

   ANSWER: B

10. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
    Q.If $\dfrac{x^2 - x + 1}{(x^2 + 1)(x^2 + x + 1)} = \dfrac{Ax + B}{x^2 + 1} + \dfrac{Cx + D}{x^2 + x + 1}$ then $A + 2B + C + 2D =$ (A) $0$ (B) $1$ (C) $-1$ (D) $2$

    ›Reveal solutionSolution

    This is a partial fractions problem where we match coefficients after clearing denominators. The value of $A+2B+C+2D$ is $0$.

    The core idea here is that when you have a rational expression and you decompose it into partial fractions, the numerators on the right-hand side are linear (since the denominators are irreducible quadratics). To find the constants, you multiply through by the common denominator and then compare coefficients of powers of $x$. That gives a system of linear equations. Once you solve for $A, B, C, D$, you just plug into the required combination.

    Let’s work through it step by step.

    1. Set up the equation We are given:

    $\frac{x^2 - x + 1}{(x^2+1)(x^2+x+1)} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+x+1}$

    Multiply both sides by $(x^2+1)(x^2+x+1)$:

    $x^2 - x + 1 = (Ax+B)(x^2+x+1) + (Cx+D)(x^2+1)$

    2. Expand both products First term:

    $(Ax+B)(x^2+x+1) = A x^3 + A x^2 + A x + B x^2 + B x + B$

    That is:

    $A x^3 + (A+B)x^2 + (A+B)x + B$

    Second term:

    $(Cx+D)(x^2+1) = C x^3 + C x + D x^2 + D$

    That is:

    $C x^3 + D x^2 + C x + D$

    3. Add them together

    $x^2 - x + 1 = (A+C)x^3 + (A+B+D)x^2 + (A+B+C)x + (B+D)$

    4. Compare coefficients Since the left side has no $x^3$ term, its coefficient is $0$. So:

    $\begin{cases} \text{Coefficient of } x^3: & A + C = 0 \quad\quad (1)\\ \text{Coefficient of } x^2: & A + B + D = 1 \quad\quad (2)\\ \text{Coefficient of } x: & A + B + C = -1 \quad\quad (3)\\ \text{Constant term:} & B + D = 1 \quad\quad (4) \end{cases}$

    5. Solve the system

       From (1): $C = -A$.

       From (4): $D = 1 - B$.

       Substitute into (2): $A + B + (1 - B) = 1 \implies A + 1 = 1 \implies A = 0$.

       Then $C = 0$ from (1).

       Now (3): $0 + B + 0 = -1 \implies B = -1$.

       Then from (4): $D = 1 - (-1) = 2$.

       So we have:

    $A = 0,\quad B = -1,\quad C = 0,\quad D = 2$

    6. Compute the required expression

    $A + 2B + C + 2D = 0 + 2(-1) + 0 + 2(2) = -2 + 4 = 2$

    Watch out

    A common mistake is to forget the constant term when expanding $(Cx+D)(x^2+1)$ — the $D$ multiplies the $1$ to give $D$, not $Dx^2$. Also, be careful with signs when comparing coefficients: the left side is $x^2 - x + 1$, so the coefficient of $x$ is $-1$, not $+1$.

    ✓Final answer

    The value is $\boxed{2}$, which corresponds to option (D).

11. TG EAPCET 2021Set eng-2021-08-06-AN1 markMCQ
    Q.The partial fraction decomposition of $\dfrac{3x+1}{(x-1)^2(x+2)}$ is (A) $\dfrac{4}{3}\dfrac{1}{(x-1)^2} + \dfrac{5}{9}\dfrac{1}{x-1} + \dfrac{5}{9}\dfrac{1}{x+2}$ (B) $-\dfrac{5}{9}\left(\dfrac{1}{x+2}\right) + \dfrac{4}{3}\dfrac{1}{(x-1)^2} + \dfrac{2}{x-1}$ (C) $-\dfrac{5}{9}\left(\dfrac{1}{x+2}\right) + \dfrac{5}{9}\dfrac{1}{x-1} + \dfrac{4}{3}\dfrac{1}{(x-1)^2}$ (D) $-\dfrac{5}{9}\left(\dfrac{1}{x+2}\right) + \dfrac{5}{9}\dfrac{1}{x-1} + \dfrac{2}{(x-1)^2}$

    ›Reveal solutionSolution

    For a repeated linear factor $(x-1)^2$, the partial fraction form must include both $\frac{A}{x-1}$ and $\frac{B}{(x-1)^2}$, plus $\frac{C}{x+2}$. Solving gives the decomposition $-\frac{5}{9}\frac{1}{x+2} + \frac{5}{9}\frac{1}{x-1} + \frac{4}{3}\frac{1}{(x-1)^2}$, which matches option (C).

    The key idea: when a denominator has a repeated linear factor like $(x-1)^2$, you cannot just write one term for it. You need a term for each power, from $1$ up to the exponent. That’s the whole reason the standard form looks the way it does.

    1. Set up the correct form. Since $(x-1)^2$ is a repeated factor, the decomposition is

    $\frac{3x+1}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}.$

    Notice: $A$ goes with the first power, $B$ with the square. A common mistake is to forget the $A/(x-1)$ term entirely — that would be wrong.

    2. Clear denominators. Multiply both sides by $(x-1)^2(x+2)$:

    $3x+1 = A(x-1)(x+2) + B(x+2) + C(x-1)^2.$

    3. Solve for the constants.

       The smartest way is to plug in convenient $x$ values that make factors zero.

       • For $B$: Set $x = 1$. Then $(x-1)=0$, so the $A$ and $C$ terms vanish:

    $3(1)+1 = B(1+2) \quad\Rightarrow\quad 4 = 3B \quad\Rightarrow\quad B = \frac{4}{3}.$

    • For $C$: Set $x = -2$. Then $(x+2)=0$, so the $A$ and $B$ terms vanish:

    $3(-2)+1 = C(-2-1)^2 \quad\Rightarrow\quad -5 = C(9) \quad\Rightarrow\quad C = -\frac{5}{9}.$

    • For $A$: Now pick any other $x$, say $x = 0$, and substitute $B$ and $C$:

    $3(0)+1 = A(0-1)(0+2) + \frac{4}{3}(0+2) + \left(-\frac{5}{9}\right)(0-1)^2.$

     That gives  
    

    $1 = A(-1)(2) + \frac{4}{3}(2) - \frac{5}{9}(1) = -2A + \frac{8}{3} - \frac{5}{9}.$

     Combine the fractions: $\frac{8}{3} = \frac{24}{9}$, so $\frac{24}{9} - \frac{5}{9} = \frac{19}{9}$.  
     Then  
    

    $1 = -2A + \frac{19}{9} \quad\Rightarrow\quad -2A = 1 - \frac{19}{9} = -\frac{10}{9} \quad\Rightarrow\quad A = \frac{5}{9}.$

    4. Write the decomposition. Putting it all together:

    $\frac{3x+1}{(x-1)^2(x+2)} = \frac{5/9}{x-1} + \frac{4/3}{(x-1)^2} - \frac{5/9}{x+2}.$

    That is exactly option (C).

    Watch out

    A very common error is to write only $\frac{B}{(x-1)^2} + \frac{C}{x+2}$, skipping the $\frac{A}{x-1}$ term. That would give a wrong decomposition because the numerator degree (1) is less than the denominator degree (3), but the repeated factor still demands a term for each power.

    Tip

    When solving for constants, always use the “root-substitution” trick first for the easiest ones ($B$ and $C$ here). Only then substitute a simple number like $0$ or $1$ for the remaining constant. It saves time and avoids solving a system of equations.

    ✓Final answer

    The correct option is (C).

12. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
    Q.If $\dfrac{3x^4 - 2x^2 + 1}{(x-2)^4} = \dfrac{A}{x-2} + \dfrac{B}{(x-2)^2} + \dfrac{C}{(x-2)^3} + \dfrac{D}{(x-2)^4}$, then $2A + 3B - C - D + E =$ (A) $0$ (B) $1$ (C) $-11$ (D) $-39$

    ›Reveal solutionSolution

    Writing the fraction in powers of $(x-2)$ gives $E=3,\ A=3,\ B=24,\ C=70,\ D=88,\ E=41$ for the five partial-fraction constants, so $2A+3B-C-D+E=-39$.

    Note on the statement: because the expression carries a constant $E$, the intended decomposition has a fifth-order denominator (five constants $A,B,C,D,E$):

    $\frac{3x^{4}-2x^{2}+1}{(x-2)^{5}}=\frac{A}{x-2}+\frac{B}{(x-2)^{2}}+\frac{C}{(x-2)^{3}}+\frac{D}{(x-2)^{4}}+\frac{E}{(x-2)^{5}}.$

    Substitute $x-2=t$ (i.e. $x=t+2$) in the numerator:

    $3(t+2)^4-2(t+2)^2+1=3t^{4}+24t^{3}+70t^{2}+88t+41.$

    Dividing by $t^{5}=(x-2)^5$ reads off the coefficients directly:

    $\frac{3t^{4}+24t^{3}+70t^{2}+88t+41}{t^{5}}=\frac{3}{t}+\frac{24}{t^{2}}+\frac{70}{t^{3}}+\frac{88}{t^{4}}+\frac{41}{t^{5}}.$

    So

    $A=3,\quad B=24,\quad C=70,\quad D=88,\quad E=41.$

    Therefore

    $2A+3B-C-D+E=2(3)+3(24)-70-88+41=6+72-70-88+41=-39.$

    ✓Final answer

    $2A+3B-C-D+E=-39$ — option (D).