Q.Integrate the following function: $\frac{2x}{(x^2+1)(x^2+3)}$

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:

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

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2. 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 $D = 1-B$. …

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3. 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 $C = 0$. …

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4. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
   Q.If $\dfrac{d}{dx} \left( \dfrac{2x+1}{(x+1)^2 (x-2)} \right) = \dfrac{A}{(x-2)^2} + \dfrac{B}{(x+2)^3} + \dfrac{C}{(x+1)^2}$ then $A+B+C=$ (A) $\dfrac{-2}{3}$ (B) $\dfrac{2}{3}$ (C) $\dfrac{1}{3}$ (D) $\dfrac{-1}{3}$

   ›Reveal solutionSolution

   The problem asks for the sum of constants in a partial-fraction-like decomposition of a derivative. By differentiating the given rational function and matching the form, we find $A+B+C = -\frac{2}{3}$, which corresponds to option (A).

   The key insight here is that the derivative of a rational function can be expressed as a sum of simpler fractions, but the given form is not a standard partial fraction decomposition — it’s a template for the derivative itself. The constants $A$, $B$, $C$ are not from splitting the original function, but from rewriting its derivative in that specific pattern. So we must differentiate first, then compare.

   Let’s work through it.

   1. Differentiate the given function.

      Let $f(x) = \dfrac{2x+1}{(x+1)^2 (x-2)}$.

      Use the quotient rule: $f'(x) = \frac{u'v - uv'}{v^2}$, where $u = 2x+1$ and $v = (x+1)^2 (x-2)$.

      First, $u' = 2$.

      For $v'$, use the product rule: $v = (x+1)^2 \cdot (x-2)$.

      Derivative: $v' = 2(x+1)(1) \cdot (x-2) + (x+1)^2 \cdot 1 = 2(x+1)(x-2) + (x+1)^2$.

      Simplify: $v' = (x+1)[2(x-2) + (x+1)] = (x+1)(2x - 4 + x + 1) = (x+1)(3x - 3) = 3(x+1)(x-1)$.

      So $f'(x) = \frac{2 \cdot (x+1)^2 (x-2) - (2x+1) \cdot 3(x+1)(x-1)}{(x+1)^4 (x-2)^2}$.

   2. Simplify the numerator.

      Factor out $(x+1)$ from the numerator (common factor):

      Numerator $= (x+1) \left[ 2(x+1)(x-2) - 3(2x+1)(x-1) \right]$.

      Expand inside:

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

      $3(2x+1)(x-1) = 3(2x^2 - 2x + x - 1) = 3(2x^2 - x - 1) = 6x^2 - 3x - 3$.

      Subtract: $(2x^2 - 2x - 4) - (6x^2 - 3x - 3) = -4x^2 + x - 1$.

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

      Thus $f'(x) = \frac{(x+1)(-4x^2 + x - 1)}{(x+1)^4 (x-2)^2} = \frac{-4x^2 + x - 1}{(x+1)^3 (x-2)^2}$.

   3. Match to the given form.

      We are told $f'(x) = \frac{A}{(x-2)^2} + \frac{B}{(x+2)^3} + \frac{C}{(x+1)^2}$.

      But our denominator has factors $(x+1)^3$ and $(x-2)^2$ — note the $(x+2)^3$ term in the given form is suspicious. That’s likely a typo in the problem statement (common in such questions); it should be $(x+1)^3$ instead of $(x+2)^3$. We’ll proceed assuming the intended form is $\frac{A}{(x-2)^2} + \frac{B}{(x+1)^3} + \frac{C}{(x+1)^2}$.

      So we need to decompose $\frac{-4x^2 + x - 1}{(x+1)^3 (x-2)^2}$ into partial fractions with denominators as above.

   4. Set up the decomposition.

      Write:

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

   Multiply both sides by $(x+1)^3 (x-2)^2$:

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

   5. Find constants by substitution.
      • Put $x = 2$: $-4(4) + 2 - 1 = -16 + 1 = -15$. …

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5. 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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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 2023Set eng-2023-05-14-FN1 markMCQ
   Q.If $\dfrac{3x+2}{(x+1)(2x^2+3)} = \dfrac{A}{x+1} + \dfrac{Bx+C}{2x^2+3}$, then $A-B+C=$ (A) 2 (B) 1 (C) 3 (D) 6

   ›Reveal solutionSolution

   We find the constants in the partial fraction decomposition by clearing denominators and equating coefficients, then compute $A - B + C = 1$.

   The problem gives a rational function and its partial fraction decomposition. The key idea: multiply both sides by the common denominator to get a polynomial identity, then match coefficients to solve for $A$, $B$, and $C$. Once we have them, the expression $A - B + C$ is straightforward.

   ---

   1. Set up the equation We have

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

   Multiply both sides by $(x+1)(2x^2+3)$ to clear denominators:

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

   2. Expand the right-hand side First term: $A(2x^2+3) = 2A x^2 + 3A$. Second term: $(Bx+C)(x+1) = Bx^2 + Bx + Cx + C = Bx^2 + (B+C)x + C$. Adding them:

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

   3. Equate coefficients Since the left side has no $x^2$ term, its coefficient is 0. So:

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

   4. Solve the system From (1): $B = -2A$. Substitute into (2): $-2A + C = 3 \Rightarrow C = 3 + 2A$. Substitute into (3): $3A + (3 + 2A) = 2 \Rightarrow 5A + 3 = 2 \Rightarrow 5A = -1 \Rightarrow A = -\frac{1}{5}$. …

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8. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
   Q.If $\dfrac{x^2 - 2}{(x^2 + 1)(x^2 + 3)} = \dfrac{Ax + B}{x^2 + 1} + \dfrac{Cx + D}{x^2 + 3}$ then D = (A) $-\dfrac{3}{2}$ (B) $-\dfrac{1}{2}$ (C) $2$ (D) $\dfrac{5}{2}$

   ›Reveal solutionSolution

   This problem involves decomposing a rational function into partial fractions with irreducible quadratic denominators. By equating coefficients after clearing denominators, we find that $D = \boxed{\frac{5}{2}}$.

   The core idea here is partial fraction decomposition, a technique used to break down complex rational functions into simpler ones. This is particularly useful in calculus for integration, but also in algebra for manipulating expressions.

   When the denominator contains irreducible quadratic factors (like $x^2+1$ or $x^2+3$, which cannot be factored into real linear terms), the corresponding numerator in the partial fraction decomposition takes the form $Ax+B$.

   In this specific problem, notice that the original numerator ($x^2-2$) contains only even powers of $x$, and the denominators ($x^2+1$, $x^2+3$) also contain only even powers of $x$. This is a strong hint that the terms with odd powers of $x$ (i.e., $Ax$ and $Cx$) in the partial fraction expansion will turn out to be zero. We will confirm this by comparing coefficients.

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

   1. Set up the equation: We are given the partial fraction decomposition:

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

   2. Clear the denominators: Multiply both sides of the equation by the common denominator $(x^2 + 1)(x^2 + 3)$:

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

   3. Expand the right-hand side: Distribute the terms on the right side:

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

   $x^2 - 2 = Ax^3 + 3Ax + Bx^2 + 3B + Cx^3 + Cx + Dx^2 + D$

   4. Group terms by powers of $x$: Rearrange the terms on the right-hand side to group coefficients of $x^3$, $x^2$, $x$, and the constant term:

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

   5. Compare coefficients:

      Now, we compare the coefficients of corresponding powers of $x$ on both sides of the equation. The left-hand side, $x^2 - 2$, can be written as $0x^3 + 1x^2 + 0x - 2$.

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

   6. Solve the system of equations for $A, B, C, D$:

      First, let's solve for $A$ and $C$ using Equations 1 and 3:

      From Equation 1, $C = -A$.

      Substitute this into Equation 3:

      $3A + (-A) = 0$

      $2A = 0$

      $A = 0$

      Since $A=0$, from $C=-A$, we get $C=0$. …

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9. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
   Q.If $\dfrac{x-2}{x^2(2x-3)} = \dfrac{A}{x} + \dfrac{B}{x^2} + \dfrac{C}{2x-3}$ then $2(A-C)=$ (A) 3B (B) 2B (C) 0 (D) B

   ›Reveal solutionSolution

   To find the coefficients A, B, and C in the partial fraction decomposition, we equate the numerators after combining the terms on the right-hand side. By substituting specific values of $x$ or comparing coefficients, we find $A = 1/9$, $B = 2/3$, and $C = -2/9$. The expression $2(A-C)$ then evaluates to $2/3$, which is equal to $B$.

   Partial fraction decomposition is a technique used to break down a complex rational function into a sum of simpler fractions. This is particularly useful in calculus for integration, but also in algebra for manipulating expressions. The core idea is that any proper rational function (where the degree of the numerator is less than the degree of the denominator) can be expressed as a sum of fractions whose denominators are the factors of the original denominator.

   When the denominator has repeated linear factors, like $x^2$ in this problem, the decomposition must include a term for each power of the factor up to its multiplicity. For a factor $(ax+b)^n$, we include terms $\frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} + \dots + \frac{A_n}{(ax+b)^n}$. For distinct linear factors, like $(2x-3)$, we simply have a term $\frac{C}{2x-3}$.

   The strategy is to combine the partial fractions on the right-hand side, equate the resulting numerator to the original numerator, and then solve for the unknown coefficients (A, B, C) by either substituting convenient values of $x$ or by comparing coefficients of like powers of $x$.

   1. Set up the equation and clear denominators: We are given the partial fraction decomposition:

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

   To find the coefficients $A$, $B$, and $C$, we first combine the terms on the right-hand side by finding a common denominator, which is $x^2(2x-3)$.
   

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

   Since the denominators are now identical, the numerators must be equal:
   

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

   Expand the right-hand side:
   

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

   Rearrange the terms by powers of $x$:
   

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

   2. Determine the coefficients using strategic substitution and comparison:

      We can find the coefficients by substituting values of $x$ that make certain terms zero, or by comparing the coefficients of $x^2$, $x$, and the constant term on both sides of the equation.

      • Find B: Substitute $x=0$ into the equation $x-2 = A(x)(2x-3) + B(2x-3) + C(x^2)$. This eliminates the terms with $A$ and $C$:

   $(0)-2 = A(0)(2(0)-3) + B(2(0)-3) + C(0)^2$

   $-2 = 0 + B(-3) + 0$

   $-2 = -3B \implies B = \frac{2}{3}$

   *   **Find C:** Substitute $x=\frac{3}{2}$ (which makes $2x-3=0$) into the equation $x-2 = A(x)(2x-3) + B(2x-3) + C(x^2)$. This eliminates the terms with $A$ and $B$:
   

   $\frac{3}{2}-2 = A\left(\frac{3}{2}\right)(0) + B(0) + C\left(\frac{3}{2}\right)^2$

   $\frac{3-4}{2} = 0 + 0 + C\left(\frac{9}{4}\right)$

   $-\frac{1}{2} = \frac{9}{4}C \implies C = -\frac{1}{2} \times \frac{4}{9} = -\frac{2}{9}$

   *   **Find A:** Now that we have $B$ and $C$, we can find $A$ by comparing the coefficients of $x^2$ from the expanded equation: …
   

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

    ›Reveal solutionSolution

    Clearing denominators and matching coefficients gives $A=0,\ B=2,\ C=0,\ D=-1$, so $A+B+C+D=1$ — option (D).

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

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

    Expand and collect powers of $x$:

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

    Match coefficients:

    • $x^3:\ A+C=0$
    • $x^2:\ B+D=1$
    • $x^1:\ 2A+C=0$
    • $x^0:\ 2B+D=3$ …

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

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12. 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 $A, B$:

       $$$$

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