Q.$\int \dfrac{x^2\,dx}{(x^2+a^2)(x^2+b^2)}$

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 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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3. 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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4. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.If $\dfrac{x^2}{2x^4 + 7x^2 + 6} = \dfrac{Ax + B}{x^2 + a} + \dfrac{Cx + D}{ax^2 + 3}$, then $A + B + C - 2D =$ (A) $2a$ (B) $-2a$ (C) $-4a$ (D) $4a$

   ›Reveal solutionSolution

   The key idea is to perform a partial fraction decomposition of the given rational expression, match coefficients, and then evaluate the required combination $A+B+C-2D$ in terms of $a$. The final result is $-4a$, so the correct option is (C).

   ---

   We start with the equation:

   $\frac{x^2}{2x^4 + 7x^2 + 6} = \frac{Ax + B}{x^2 + a} + \frac{Cx + D}{ax^2 + 3}$

   The denominator on the left factors as a quadratic in $x^2$. Let’s see why this approach works: partial fractions decompose a complicated rational expression into simpler pieces whose numerators are linear (since denominators are irreducible quadratics). The constants $a$ is not a variable to solve for—it’s a parameter that appears in the factorization. Our job is to find $A, B, C, D$ in terms of $a$, then compute the given combination.

   ---

   1. Factor the denominator on the left.

   Let $y = x^2$. Then:

   $2x^4 + 7x^2 + 6 = 2y^2 + 7y + 6$

   Factor the quadratic in $y$:

   $2y^2 + 7y + 6 = (2y + 3)(y + 2)$

   Substitute back $y = x^2$:

   $2x^4 + 7x^2 + 6 = (2x^2 + 3)(x^2 + 2)$

   So the left-hand side becomes:

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

   ---

   2. Match the denominators with the right-hand side.

   The right-hand side has denominators $x^2 + a$ and $ax^2 + 3$. For the decomposition to match, we must have:

   $x^2 + a = x^2 + 2 \quad \Rightarrow \quad a = 2$

   and

   $ax^2 + 3 = 2x^2 + 3 \quad \Rightarrow \quad a = 2$

   Both conditions give $a = 2$. So the parameter $a$ is actually determined by the factorization. This is a crucial observation.

   Watch out

   A common mistake is to treat $a$ as an unknown constant to be solved from the equation after clearing denominators. But here $a$ is already fixed by the factorization—it’s not free. The equation holds for all $x$ only when $a = 2$.

   ---

   3. Set up the partial fractions with $a = 2$.

   We have:

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

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

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

   ---

   4. Expand and collect like terms.

   First term:

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

   Second term:

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

   Sum:

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

   ---

   5. Equate coefficients. …

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5. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.If $\dfrac{x+3}{(1-x)^2(1+x^2)} = \dfrac{A}{(1-x)} + \dfrac{B}{(1-x)^2} + \dfrac{Cx+D}{2(1+x^2)}$ then $B^2 + C^2 + D^2 =$ (A) $\dfrac{3}{4}$ (B) $4 \dfrac{1}{2}$ (C) $14$ (D) $4$

   ›Reveal solutionSolution

   We solve for the partial fraction coefficients by clearing denominators and equating numerators, then compute $B^2 + C^2 + D^2$ to find the result is $\frac{3}{4}$, which corresponds to option (A).

   The problem gives a partial fraction decomposition of a rational function. The key idea is to multiply both sides by the common denominator to obtain a polynomial identity, then solve for the unknown constants $A$, $B$, $C$, $D$ by comparing coefficients or substituting convenient values of $x$. Once we have $B$, $C$, $D$, we compute the sum of their squares.

   1. Set up the equation We have

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

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

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

   2. Clear the fraction in the last term Multiply the entire equation by 2 to avoid fractions:

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

   So:

   $2x+6 = 2A(1-x)(1+x^2) + 2B(1+x^2) + (Cx+D)(1-x)^2.$

   3. Expand each term

      • First term: $2A(1-x)(1+x^2) = 2A[(1)(1+x^2) - x(1+x^2)] = 2A(1+x^2 - x - x^3) = 2A(-x^3 - x + 1 + x^2)$. Better to expand systematically: $(1-x)(1+x^2) = 1 + x^2 - x - x^3$. So $2A(1 - x + x^2 - x^3)$.
      • Second term: $2B(1+x^2) = 2B + 2B x^2$.
      • Third term: $(Cx+D)(1-x)^2 = (Cx+D)(1 - 2x + x^2)$. Expand: $Cx(1 - 2x + x^2) = Cx - 2Cx^2 + Cx^3$ $D(1 - 2x + x^2) = D - 2Dx + Dx^2$ Sum: $Cx^3 + (-2C + D)x^2 + (C - 2D)x + D$.

   4. Collect coefficients

      The left side is $2x + 6$, which is $0\cdot x^3 + 0\cdot x^2 + 2x + 6$.

      The right side, collecting powers:

      • $x^3$: from first term: $2A(-1) = -2A$; from third: $C$. So coefficient: $C - 2A$.
      • $x^2$: from first: $2A(1) = 2A$; from second: $2B$; from third: $(-2C + D)$. So: $2A + 2B - 2C + D$.
      • $x^1$: from first: $2A(-1) = -2A$; from third: $(C - 2D)$. So: $-2A + C - 2D$.
      • Constant: from first: $2A(1) = 2A$; from second: $2B$; from third: $D$. So: $2A + 2B + D$.

      Equate to $0x^3 + 0x^2 + 2x + 6$:

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

   5. Solve the system From (1): $C = 2A$. …

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6. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
   Q.If $\dfrac{6x^4 + 13x^3 + 2x^2 - x + 3}{2x^2 + 3x - 2} = f(x) + \dfrac{A}{ax-1} + \dfrac{B}{x+b}$, then $f(1) + a \cdot B + b \cdot A =$ (A) 8 (B) 12 (C) 4 (D) 6

   ›Reveal solutionSolution

   Long division gives $f(x)=3x^2+2x+1$ with $a=2,b=2,A=2,B=-1$, so $f(1)+aB+bA = 6-2+4 = 8$ (option A).

   Factor the denominator: $2x^2+3x-2 = (2x-1)(x+2)$, so $ax-1 = 2x-1 \Rightarrow a=2$ and $x+b = x+2 \Rightarrow b=2$.

   Polynomial division of $6x^4+13x^3+2x^2-x+3$ by $2x^2+3x-2$:

   $f(x) = 3x^2 + 2x + 1, \qquad \text{remainder } = 5.$

   Partial fractions of the remainder:

   $\frac{5}{(2x-1)(x+2)} = \frac{A}{2x-1} + \frac{B}{x+2} \ \Rightarrow\ 5 = A(x+2) + B(2x-1).$ …

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7. 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: $B + D = 1$ …

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8. 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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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 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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11. 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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12. 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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