Q.Integrate the following function: $\frac{5x}{(x + 1)(x^2 - 4)}$

Showing the 12 most recent of 33 on this concept.

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

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2. 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:

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

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3. 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  
    $$ …
   

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4. 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 $B = -A = \frac{17}{5}$. …

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5. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
   Q.If $\dfrac{x^2+7}{(x^2+1)(x-2)} = \dfrac{A}{x-2} + \dfrac{Bx+C}{x^2+1}$, then the determinant of the matrix $\begin{pmatrix} A & B \\ C & \dfrac{2}{5} \end{pmatrix}$ is (A) $5$ (B) $-5$ (C) $\dfrac{94}{25}$ (D) $-2$

   ›Reveal solutionSolution

   We decompose the given rational function into partial fractions, solve for $A$, $B$, and $C$, then compute the determinant of the $2\times2$ matrix formed by them. The determinant is $\boxed{-5}$.

   The problem gives a partial fraction decomposition and asks for the determinant of a matrix built from the coefficients. The key is to find $A$, $B$, and $C$ correctly — then the determinant is just a quick calculation.

   We start with the identity:

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

   This holds for all $x$ (except where denominators vanish). Multiply both sides by $(x^2+1)(x-2)$ to clear denominators:

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

   1. Expand the right-hand side

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

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

   Adding them:

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

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

   $\begin{cases} A + B = 1 & \text{(coefficient of $x^2$)} \\ C - 2B = 0 & \text{(coefficient of $x$)} \\ A - 2C = 7 & \text{(constant term)} \end{cases}$

   3. Solve the system From the second equation: $C = 2B$. Substitute into the third: $A - 2(2B) = A - 4B = 7$. From the first: $A = 1 - B$. Plug into $A - 4B = 7$:

   $(1 - B) - 4B = 7 \quad\Rightarrow\quad 1 - 5B = 7 \quad\Rightarrow\quad -5B = 6 \quad\Rightarrow\quad B = -\frac{6}{5}$

   Then $A = 1 - (-\frac{6}{5}) = 1 + \frac{6}{5} = \frac{11}{5}$.

   And $C = 2B = 2 \cdot (-\frac{6}{5}) = -\frac{12}{5}$.

   So we have:

   $A = \frac{11}{5},\quad B = -\frac{6}{5},\quad C = -\frac{12}{5}$

   4. Form the matrix and compute the determinant The matrix is: $$$$

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6. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.If $\dfrac{x^2-3}{(x+2)(x^2+1)} = \dfrac{A}{x+2} + \dfrac{Bx+C}{x^2+1}$ then $3A+2B-C=$ (A) $\dfrac{8}{5}$ (B) $\dfrac{16}{5}$ (C) $\dfrac{3}{5}$ (D) $\dfrac{19}{5}$

   ›Reveal solutionSolution

   Solving the partial-fraction system gives $A=\tfrac15,\ B=\tfrac45,\ C=-\tfrac85$, so $3A+2B-C=\tfrac{19}{5}$, option (D).

   Clear denominators in $\dfrac{x^2-3}{(x+2)(x^2+1)}=\dfrac{A}{x+2}+\dfrac{Bx+C}{x^2+1}$:

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

   Matching coefficients:

   $A+B=1,\qquad 2B+C=0,\qquad A+2C=-3.$ …

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7. 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: …

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8. TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
   Q.If $\dfrac{x^2 - 3x + 2}{(x-4)(x-3)^2} = \dfrac{A}{x-4} + \dfrac{B}{x-3} + \dfrac{C}{(x-3)^2}$ then $A+B+C=$ (A) $1$ (B) $0$ (C) $-1$ (D) $5$

   ›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 is $A+B+C = 0$.

   We are given the partial fraction decomposition:

   $\frac{x^2 - 3x + 2}{(x-4)(x-3)^2} = \frac{A}{x-4} + \frac{B}{x-3} + \frac{C}{(x-3)^2}$

   We need $A+B+C$. Instead of solving for each constant individually and then adding, we can find the sum directly by cleverly evaluating the equality at a convenient $x$.

   ---

   Concept & Intuition

   When two rational expressions are equal for all $x$ (except where denominators vanish), their numerators are equal after clearing denominators. If we multiply both sides by $(x-4)(x-3)^2$, we get a polynomial identity. Then, to find $A+B+C$, we can plug in a value of $x$ that makes the coefficients combine nicely — here $x=2$ works because it zeroes out the original numerator and simplifies the right-hand side.

   ---

   Step-by-step solution

   1. Clear denominators Multiply both sides by $(x-4)(x-3)^2$:

   $x^2 - 3x + 2 = A(x-3)^2 + B(x-4)(x-3) + C(x-4)$

   This holds for all $x$ (except the poles, but as polynomials they agree everywhere).

   2. Choose a clever $x$ to get a relation among $A, B, C$ We want $A+B+C$. Notice that if we set $x=2$, the left-hand side becomes:

   $2^2 - 3(2) + 2 = 4 - 6 + 2 = 0$

   On the right-hand side:

   $A(2-3)^2 + B(2-4)(2-3) + C(2-4) = A(1) + B(-2)(-1) + C(-2)$

   Simplify:

   $= A + 2B - 2C$

   So we have:

   $A + 2B - 2C = 0 \quad \text{(Equation 1)}$

   3. Find another relation

      To get $A+B+C$, we need one more equation. A natural choice is to set $x=0$:

      Left-hand side: $0^2 - 0 + 2 = 2$

      Right-hand side: $A(0-3)^2 + B(0-4)(0-3) + C(0-4) = 9A + 12B - 4C$

      So:

   $9A + 12B - 4C = 2 \quad \text{(Equation 2)}$

   4. Combine to find $A+B+C$ We want $S = A+B+C$. Notice that Equation 1 is $A + 2B - 2C = 0$. If we subtract $S$ from something? Better: Let’s express $C$ in terms of $A$ and $B$ from Equation 1:

   $A + 2B = 2C \quad \Rightarrow \quad C = \frac{A + 2B}{2}$

   Then $S = A + B + \frac{A+2B}{2} = \frac{2A + 2B + A + 2B}{2} = \frac{3A + 4B}{2}$.

   Now use Equation 2: $9A + 12B - 4\left(\frac{A+2B}{2}\right) = 2$

   Simplify: $9A + 12B - 2(A+2B) = 2$

   $\Rightarrow 9A + 12B - 2A - 4B = 2$

   $\Rightarrow 7A + 8B = 2$

   We have two equations in $A$ and $B$:

   $\begin{cases} A + 2B = 2C \quad \text{(already used)} \\ 7A + 8B = 2 \end{cases}$

   But we don’t actually need $A$ and $B$ separately — we need $S = \frac{3A+4B}{2}$. Notice $7A+8B = 2$ is almost $2(3A+4B)$? No: $2(3A+4B) = 6A+8B$, not $7A+8B$. So we need one more step.

   5. Alternative: Direct evaluation at $x=1$

      Set $x=1$:

      LHS: $1 - 3 + 2 = 0$

      RHS: $A(1-3)^2 + B(1-4)(1-3) + C(1-4) = A(4) + B(-3)(-2) + C(-3) = 4A + 6B - 3C$

      So:

   $4A + 6B - 3C = 0 \quad \text{(Equation 3)}$

   Now we have three equations:

   $\begin{cases} A + 2B - 2C = 0 \quad (1) \\ 9A + 12B - 4C = 2 \quad (2) \\ 4A + 6B - 3C = 0 \quad (3) \end{cases}$

   Subtract (3) from (2): $(9A-4A) + (12B-6B) + (-4C+3C) = 2$

   $\Rightarrow 5A + 6B - C = 2$ (Equation 4) …

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9. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.If $\dfrac{x^2+1}{(x^4+5x^2+6)(x^6+x^4)} = \dfrac{A}{x^4} + \dfrac{B}{x^2} + \dfrac{C}{x^2+2} + \dfrac{D}{x^2+3}$, then $A-B=$ (A) $\dfrac{13}{36}$ (B) $\dfrac{11}{36}$ (C) $\dfrac{2}{9}$ (D) $-\dfrac{1}{2}$

   ›Reveal solutionSolution

   The key is to factor the denominator completely, then match the given partial-fraction form to the actual decomposition. After simplifying, we find $A - B = \frac{11}{36}$, which is option (B).

   The problem gives a partial-fraction expansion with four terms, but the left-hand side has a denominator that factors into products of quadratics and powers of $x$. The trick is that the given form is not the standard partial-fraction decomposition — it’s a specific rearrangement. We need to find $A$ and $B$ by comparing coefficients after clearing denominators.

   Let’s work through it step by step.

   1. Factor the denominator completely. The left-hand side is

   $\frac{x^2+1}{(x^4+5x^2+6)(x^6+x^4)}.$

   First, $x^6 + x^4 = x^4(x^2+1)$.

   Next, $x^4+5x^2+6$ is quadratic in $x^2$: let $u = x^2$, then $u^2+5u+6 = (u+2)(u+3) = (x^2+2)(x^2+3)$.

   So the whole denominator is

   $(x^2+2)(x^2+3) \cdot x^4 (x^2+1).$

   Notice the $x^2+1$ in the numerator cancels with the $x^2+1$ in the denominator!

   Hence the expression simplifies to

   $\frac{1}{x^4 (x^2+2)(x^2+3)}.$

   2. Set up the given partial-fraction form. We are told

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

   Multiply both sides by $x^4 (x^2+2)(x^2+3)$ to clear denominators:

   $1 = A (x^2+2)(x^2+3) + B x^2 (x^2+2)(x^2+3) + C x^4 (x^2+3) + D x^4 (x^2+2).$

   3. Expand and collect powers of $x$.

      Compute each term:

      • $A(x^2+2)(x^2+3) = A(x^4 + 5x^2 + 6)$.
      • $B x^2 (x^4 + 5x^2 + 6) = B(x^6 + 5x^4 + 6x^2)$.
      • $C x^4 (x^2+3) = C(x^6 + 3x^4)$.
      • $D x^4 (x^2+2) = D(x^6 + 2x^4)$.

      Summing, the coefficient of each power of $x$ on the right must match the left side, which is just the constant $1$ (i.e., coefficient of $x^0$ is 1, all others 0).

      Collect by powers:

      • $x^6$: $B + C + D = 0$
      • $x^4$: $A + 5B + 3C + 2D = 0$ …

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10. 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: …
    

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11. TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
    Q.If $\dfrac{9x-7}{(x+3)(x^2+1)} = \dfrac{A}{x+3} + \dfrac{Bx+C}{x^2+1}$ where $A, B, C \in \mathbb{R}$, then $A+B+C=$ (A) $\dfrac{17}{5}$ (B) $\dfrac{-6}{5}$ (C) $\dfrac{6}{5}$ (D) $\dfrac{-17}{5}$

    ›Reveal solutionSolution

    Use partial fractions to match coefficients; solving gives $A = -\frac{17}{5}$, $B = \frac{17}{5}$, $C = -\frac{6}{5}$, so $A+B+C = -\frac{6}{5}$.

    The core idea here is partial fraction decomposition — breaking a rational function into simpler pieces that are easier to integrate or manipulate. The given form tells us exactly what denominators to expect: a linear factor $(x+3)$ and an irreducible quadratic $(x^2+1)$. The numerator over the quadratic is linear ($Bx+C$) because the quadratic can't factor further over the reals.

    We don't integrate here; we just find the constants $A$, $B$, $C$ by equating numerators after clearing denominators.

    1. Clear denominators. 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:

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

    3. Equate coefficients with the left-hand side $9x - 7$, which is $0x^2 + 9x - 7$:

       • Coefficient of $x^2$: $A + B = 0$ → $B = -A$.
       • Coefficient of $x$: $3B + C = 9$.
       • Constant term: $A + 3C = -7$.

    4. Solve the system. From $B = -A$, substitute into $3B + C = 9$:

    $3(-A) + C = 9 \quad\Rightarrow\quad -3A + C = 9.$

    Now we have:

    $-3A + C = 9,$

    $A + 3C = -7.$

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

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12. TG EAPCET 2021Set eng-2021-08-06-FN1 markMCQ
    Q.If $\dfrac{32x^2 + 186x}{(x^2+1)(x+5)} = \dfrac{37x+1}{x^2+1} + \dfrac{\lambda}{x+5}$, then $\dfrac{\lambda}{2} =$ (A) $-5$ (B) $-\dfrac{7}{2}$ (C) $-\dfrac{3}{2}$ (D) $-\dfrac{5}{2}$

    ›Reveal solutionSolution

    To find $\lambda$, we combine the fractions on the right side, equate the numerators, and then substitute a strategic value for $x$ to solve for $\lambda$. The value of $\lambda$ is $-5$, so $\frac{\lambda}{2}$ is $\boxed{-\frac{5}{2}}$.

    The problem asks us to find the value of $\frac{\lambda}{2}$ given an identity involving partial fractions. The core idea here is that if two rational expressions are equal, and their denominators are identical, then their numerators must also be identical. This allows us to form an equation involving $x$ and $\lambda$, which we can then solve for $\lambda$.

    There are two main approaches to solve for the unknown constants in partial fraction decomposition:

    1. Equating Coefficients: Expand the right-hand side, collect terms by powers of $x$, and then equate the coefficients of corresponding powers of $x$ on both sides of the equation. This will give a system of linear equations to solve for the constants.
    2. Substitution Method: Choose specific values of $x$ that simplify the equation, often by making certain terms zero. This is particularly effective when dealing with linear factors in the denominator.

    In this problem, the substitution method is more direct for finding $\lambda$.

    Here's how we solve it step-by-step:

    1. Combine the fractions on the right-hand side: The given equation is:

    $\frac{32x^2 + 186x}{(x^2+1)(x+5)} = \frac{37x+1}{x^2+1} + \frac{\lambda}{x+5}$

    To combine the terms on the right-hand side, we find a common denominator, which is $(x^2+1)(x+5)$:
    

    $\frac{37x+1}{x^2+1} + \frac{\lambda}{x+5} = \frac{(37x+1)(x+5)}{(x^2+1)(x+5)} + \frac{\lambda(x^2+1)}{(x+5)(x^2+1)}$

    $= \frac{(37x+1)(x+5) + \lambda(x^2+1)}{(x^2+1)(x+5)}$

    2. Equate the numerators: Since the denominators on both sides of the original equation are now identical, the numerators must be equal for the identity to hold true for all values of $x$:

    $32x^2 + 186x = (37x+1)(x+5) + \lambda(x^2+1)$

    3. Solve for $\lambda$ using a strategic substitution: …

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