Q.Integrate the following function: $\int \frac{dx}{x(x^2+1)}$ equals (A) $\log |x| - \frac{1}{2} \log (x^2+1) + \text{C}$ (B) $\log |x| + \frac{1}{2} \log (x^2+1) + \text{C}$ (C) $-\log |x| + \frac{1}{2} \log (x^2+1) + \text{C}$ (D) $\frac{1}{2} \log |x| + \log (x^2+1) + \text{C}$

Showing the 12 most recent of 63 on this concept.

1. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{x^2}{(x^2-1)(x^2+1)}dx=$ (A) $\dfrac{1}{4}\log\left|\dfrac{x+1}{x-1}\right|-\dfrac{1}{2}\mathrm{Tan}^{-1}x+c$ (B) $\dfrac{1}{4}\log\left|\dfrac{x-1}{x+1}\right|+\dfrac{1}{2}\mathrm{Tan}^{-1}x+c$ (C) $\dfrac{1}{4}\log\left|\dfrac{x-1}{x+1}\right|-\dfrac{1}{2}\mathrm{Tan}^{-1}x+c$ (D) $\dfrac{1}{4}\log\left|\dfrac{x+1}{x-1}\right|+\dfrac{1}{2}\mathrm{Tan}^{-1}x+c$

   ›Reveal solutionSolution

   Splitting $\dfrac{x^2}{x^4-1}$ into a sum of $\dfrac{1}{x^2-1}$ and $\dfrac{1}{x^2+1}$ (halved) gives a standard log + arctan combination.

   Concept and Intuition

   Rather than doing full partial fractions with four unknowns, it's faster to notice $x^4-1=(x^2-1)(x^2+1)$ and that $\dfrac{1}{x^2-1}+\dfrac{1}{x^2+1}=\dfrac{(x^2+1)+(x^2-1)}{x^4-1}=\dfrac{2x^2}{x^4-1}$. This directly gives $\dfrac{x^2}{x^4-1}$ as half that sum — a shortcut avoiding solving for four separate constants.

   Step-by-Step Solution

   1. Write the denominator as $x^4-1=(x^2-1)(x^2+1)$.
   2. Observe: $\dfrac{1}{x^2-1}+\dfrac{1}{x^2+1}=\dfrac{2x^2}{x^4-1}$, so $\dfrac{x^2}{x^4-1}=\dfrac12\left[\dfrac{1}{x^2-1}+\dfrac{1}{x^2+1}\right]$.
   3. Use the standard integrals: $\displaystyle\int\dfrac{dx}{x^2-1}=\dfrac12\log\left|\dfrac{x-1}{x+1}\right|+c_1$ and $\displaystyle\int\dfrac{dx}{x^2+1}=\tan^{-1}x+c_2$.
   4. Combine: $\displaystyle\int\dfrac{x^2}{x^4-1}dx=\dfrac12\left[\dfrac12\log\left|\dfrac{x-1}{x+1}\right|+\tan^{-1}x\right]+c=\dfrac14\log\left|\dfrac{x-1}{x+1}\right|+\dfrac12\tan^{-1}x+c$.

   Common Mistakes …

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2. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{x+1}{x^3-1}\,dx=$ (A) $\dfrac{1}{3}\log\left(\dfrac{x+1}{x^2+x+1}\right)+c$ (B) $\dfrac{1}{3}\log\left(\dfrac{(x-1)^2}{x^2+x+1}\right)+c$ (C) $\dfrac{1}{3}\log\left(\dfrac{x-1}{x^2+x+1}\right)+c$ (D) $\dfrac{1}{3}\log\left(\dfrac{(x+1)^2}{x^2-x+1}\right)+c$

   ›Reveal solutionSolution

   This is a rational-function integral solved by partial fractions after factoring $x^3-1$; the result combines into a single log of $\dfrac{(x-1)^2}{x^2+x+1}$.

   Concept and Intuition

   Factor the cubic denominator as difference of cubes, then split into a linear-factor term (giving a plain log) and an irreducible-quadratic term (giving a log plus, here, no arctangent term since the numerator works out to be an exact multiple of the quadratic's derivative).

   Step-by-Step Solution

   1. $x^3-1=(x-1)(x^2+x+1)$.
   2. Write $\dfrac{x+1}{(x-1)(x^2+x+1)}=\dfrac{A}{x-1}+\dfrac{Bx+C}{x^2+x+1}$.
   3. $x+1=A(x^2+x+1)+(Bx+C)(x-1)$. At $x=1$: $2=3A\Rightarrow A=\frac23$.
   4. Matching $x^2$: $A+B=0\Rightarrow B=-\frac23$. Matching constants: $A-C=1\Rightarrow C=-\frac13$.
   5. $\displaystyle\int\frac{2/3}{x-1}dx=\frac23\log|x-1|$.
   6. $\displaystyle\int\frac{-\frac23x-\frac13}{x^2+x+1}dx=-\frac13\int\frac{2x+1}{x^2+x+1}dx=-\frac13\log(x^2+x+1)$ (numerator is exactly the derivative of the denominator). …

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3. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
   Q.$\int \dfrac{x-1}{(x-2)(x-3)}\,dx=$ (A) $2\log|x-3|+\log|x-2|+c$ (B) $\log|x-3|-\log|x-2|+c$ (C) $\log|x-3|^2-\log|x+2|+c$ (D) $\log\left|\dfrac{(x-3)^2}{x-2}\right|+c$

   ›Reveal solutionSolution

   A rational function with distinct linear factors in the denominator — resolve into partial fractions, integrate each term as a log, then combine using log rules.

   Concept and Intuition

   Any proper rational function with distinct linear denominator factors can be split into simple fractions $\frac{A}{x-a}+\frac{B}{x-b}$, each of which integrates to $A\log|x-a|$. Combining the two logs at the end into a single log of a ratio/power lets you match against answer choices written as one combined logarithm.

   Step-by-Step Solution

   1. Write $\dfrac{x-1}{(x-2)(x-3)} = \dfrac{A}{x-2}+\dfrac{B}{x-3}$, so $x-1 = A(x-3)+B(x-2)$.
   2. Put $x=2$: $2-1 = A(2-3) \Rightarrow 1 = -A \Rightarrow A=-1$.
   3. Put $x=3$: $3-1 = B(3-2) \Rightarrow 2=B$.
   4. So $\displaystyle\int\frac{x-1}{(x-2)(x-3)}dx = -\log|x-2| + 2\log|x-3| + c$. …

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4. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.$\int \frac{dx}{\sin x + \sin 2x} =$ (A) $\frac{1}{4}\log|1-\cos x| + \frac{1}{3}\log|1+\cos x| - \frac{2}{3}\log|1+\cos 2x| + c$ (B) $\frac{1}{3}\log|1-\cos x| + \frac{1}{4}\log|1+\cos x| + \frac{1}{3}\log|1+\cos 2x| + c$ (C) $\frac{1}{6}\log|1-\cos x| + \frac{1}{2}\log|1+\cos x| - \frac{2}{3}\log|1+2\cos x| + c$ (D) $\frac{1}{6}\log|1-\cos x| + \frac{1}{4}\log|1+\cos x| + \frac{2}{3}\log|1+2\cos x| + c$

   ›Reveal solutionSolution

   Factoring $\sin x+\sin2x=\sin x(1+2\cos x)$ and substituting $t=\cos x$ reduces this to a rational-function partial-fractions integral, giving $\frac16\log|1-\cos x|+\frac12\log|1+\cos x|-\frac23\log|1+2\cos x|+c$.

   Concept and Intuition

   Whenever an integral has $\sin x$ (an odd power effectively) times other cosine factors in the denominator, multiplying numerator and denominator by $\sin x$ turns $\sin^2x$ into $1-\cos^2x$, which lets us substitute $t=\cos x$ and reduce the whole problem to partial fractions of a rational function in $t$ — a completely mechanical final step.

   Step-by-Step Solution

   1. Factor the denominator: $\sin x+\sin2x = \sin x + 2\sin x\cos x = \sin x(1+2\cos x)$.
   2. So the integral is $\displaystyle\int\frac{dx}{\sin x(1+2\cos x)}$.
   3. Multiply top and bottom by $\sin x$: $\displaystyle\int\frac{\sin x\,dx}{\sin^2x(1+2\cos x)}=\int\frac{\sin x\,dx}{(1-\cos^2x)(1+2\cos x)}=\int\frac{\sin x\,dx}{(1-\cos x)(1+\cos x)(1+2\cos x)}$.
   4. Substitute $t=\cos x$, $dt=-\sin x\,dx$: integral $=-\displaystyle\int\frac{dt}{(1-t)(1+t)(1+2t)}$.
   5. Partial fractions: $\dfrac{1}{(1-t)(1+t)(1+2t)}=\dfrac{A}{1-t}+\dfrac{B}{1+t}+\dfrac{C}{1+2t}$. Evaluating at $t=1$: $A=\tfrac16$. At $t=-1$: $B=\tfrac12$. At $t=-\tfrac12$: $C=-\tfrac23$. …

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5. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{x}{x^2 - 5x + 4}\,dx =$ (A) $\dfrac{1}{3}\log\dfrac{(x-4)^4}{|x-1|} + c$ (B) $\dfrac{4}{3}\log\dfrac{|x-4|}{(x-1)^4} + c$ (C) $-\dfrac{1}{3}\log\dfrac{(x-4)^2}{|x-1|}$ (D) $-\dfrac{4}{3}\log\dfrac{|x-4|}{(x-1)^4} + c$

   ›Reveal solutionSolution

   Partial fraction decomposition of a rational function with distinct linear factors, followed by direct log integration and recombination, gives $\dfrac13\log\dfrac{(x-4)^4}{|x-1|}+c$.

   Concept and Intuition

   Whenever the denominator of a rational integrand factors into distinct linear terms, partial fractions break it into simpler pieces, each of which integrates to a logarithm. Combining the two resulting logarithm terms back into a single log-of-a-ratio (using log rules $a\log m - b\log n = \log\frac{m^a}{n^b}$) is what makes the answer match a compact multiple-choice form.

   Step-by-Step Solution

   1. Factor the denominator: $x^2-5x+4 = (x-1)(x-4)$.
   2. Write $\dfrac{x}{(x-1)(x-4)} = \dfrac{A}{x-1}+\dfrac{B}{x-4}$.
   3. Multiply through: $x = A(x-4)+B(x-1)$.
   4. Set $x=1$: $1 = A(1-4) = -3A \Rightarrow A = -\dfrac13$.
   5. Set $x=4$: $4 = B(4-1) = 3B \Rightarrow B = \dfrac43$.
   6. So the integral is $\displaystyle\int\left(-\dfrac{1/3}{x-1}+\dfrac{4/3}{x-4}\right)dx = -\dfrac13\log|x-1| + \dfrac43\log|x-4| + c$. …

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6. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.$\int\dfrac{2x^2-3}{(x^2-4)(x^2+1)}dx=A\tan^{-1}x+B\log(x-2)+C\log(x+2)$ then $6A+7B-5C=$ (A) $9$ (B) $10$ (C) $6$ (D) $8$

   ›Reveal solutionSolution

   Partial fractions of a rational function whose denominator has both real linear factors and an irreducible quadratic factor; the required integral form pins down the decomposition.

   Concept and Intuition

   The target antiderivative form $A\tan^{-1}x+B\log(x-2)+C\log(x+2)$ tells us exactly what partial-fraction decomposition must have produced it: a term $A/(x^2+1)$ (integrates to $A\tan^{-1}x$), and terms $B/(x-2)$, $C/(x+2)$.

   Step-by-Step Solution

   1. Write $\dfrac{2x^2-3}{(x-2)(x+2)(x^2+1)}=\dfrac{A}{x^2+1}+\dfrac{B}{x-2}+\dfrac{C}{x+2}$.
   2. Multiply through: $2x^2-3=A(x^2-4)+B(x+2)(x^2+1)+C(x-2)(x^2+1)$.
   3. Set $x=2$: $5=A(0)+B(4)(5)+0\Rightarrow 20B=5\Rightarrow B=\tfrac14$.
   4. Set $x=-2$: $5=0+0+C(-4)(5)\Rightarrow -20C=5\Rightarrow C=-\tfrac14$. …

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7. AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
   Q.If $\dfrac{x}{(x-1)(x^2+1)^2} = \dfrac{1}{4}\left[\dfrac{1}{x-1} - \dfrac{x+1}{x^2+1}\right]+y$, then $y =$ (A) $\dfrac{1}{2}\left[\dfrac{1-x}{(x^2+1)^2}\right]$ (B) $\dfrac{1+x}{3(x^2+1)^2}$ (C) $\dfrac{1-x}{(x^2-1)^2}$ (D) $\dfrac{1+x}{(x^2+1)^2}$

   ›Reveal solutionSolution

   This tests partial fraction decomposition with a repeated irreducible quadratic factor, then matching the remaining unaccounted term to $y$.

   Concept and Intuition

   For a denominator $(x-1)(x^2+1)^2$, the full decomposition has the form $\dfrac{A}{x-1}+\dfrac{Bx+C}{x^2+1}+\dfrac{Dx+E}{(x^2+1)^2}$. The problem already gives the first two pieces combined as $\tfrac14\left[\tfrac{1}{x-1}-\tfrac{x+1}{x^2+1}\right]$, so $y$ must be exactly the third piece.

   Step-by-Step Solution

   1. Write $\dfrac{x}{(x-1)(x^2+1)^2} = \dfrac{A}{x-1}+\dfrac{Bx+C}{x^2+1}+\dfrac{Dx+E}{(x^2+1)^2}$.
   2. Multiply through: $x = A(x^2+1)^2 + (Bx+C)(x-1)(x^2+1) + (Dx+E)(x-1)$.
   3. Set $x=1$: $1 = 4A \Rightarrow A=\tfrac14$.
   4. Expand and match coefficients of $x^4,x^3,x^2,x^1,x^0$: this yields $B=-\tfrac14,\ C=-\tfrac14,\ D=-\tfrac12,\ E=\tfrac12$.
   5. So the full decomposition is $\dfrac{1/4}{x-1} - \dfrac14\cdot\dfrac{x+1}{x^2+1} + \dfrac{(1-x)/2}{(x^2+1)^2}$. …

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8. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.If $\dfrac{x}{(x^2+1)^2(x-1)} = \dfrac{Ax+B}{x^2+1} + \dfrac{Cx+D}{(x^2+1)^2} + \dfrac{E}{x-1}$, then $A + B - C + 2D =$ (A) $\dfrac{1}{2}$ (B) $1$ (C) $\dfrac{3}{2}$ (D) $2$

   ›Reveal solutionSolution

   This tests standard partial-fraction decomposition with a repeated irreducible quadratic factor, using a mix of "plug in a root" and "match coefficients" techniques. The final computed value is $A+B-C+2D=1$.

   Concept and Intuition

   When the denominator has an irreducible quadratic factor repeated twice, $(x^2+1)^2$, along with a simple linear factor $(x-1)$, the partial fraction form needs a linear numerator ($Ax+B$, $Cx+D$) over each power of the quadratic, plus a constant ($E$) over the linear factor. The cleanest way to solve is: clear denominators, plug in the linear factor's root to isolate $E$ instantly, then expand the rest and match coefficients of each power of $x$ to get the remaining unknowns.

   Step-by-Step Solution

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

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

   2. Find $E$ quickly: set $x=1$. The first two terms vanish (each has a factor of $(x-1)$), leaving $1 = E(1^2+1)^2 = 4E$, so $E=\dfrac14$.
   3. Expand the rest. First, $(x^2+1)(x-1) = x^3-x^2+x-1$, so

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

   Next, $(Cx+D)(x-1) = Cx^2+(D-C)x-D$. And $E(x^2+1)^2 = Ex^4+2Ex^2+E$.

   4. Collect coefficients by power of $x$ and equate to the right-hand side of the original equation (which is just $x$, so coefficients are $0,0,0,1,0$ for $x^4,x^3,x^2,x^1,x^0$ respectively):

   • $x^4$: $A+E=0 \Rightarrow A=-E=-\dfrac14$.
   • $x^3$: $-A+B=0 \Rightarrow B=A=-\dfrac14$.
   • $x^2$: $(A-B)+C+2E=0$. Since $A=B$, this gives $C=-2E=-\dfrac12$. …

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9. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
   Q.$\frac{2x^2+1}{x^3-1} = \frac{A}{x-1} + \frac{Bx+C}{x^2+x+1} \Rightarrow 7A+2B+C =$ (A) $8$ (B) $9$ (C) $10$ (D) $11$

   ›Reveal solutionSolution

   Clear denominators in the partial fraction decomposition, use $x=1$ to find $A$ quickly, then match coefficients for $B$ and $C$.

   Concept and Intuition

   Since $x^3-1=(x-1)(x^2+x+1)$, the partial fraction form given is standard. Plugging in the root of the linear factor ($x=1$) isolates $A$ immediately; matching remaining coefficients (or plugging in more values) gives $B,C$.

   Step-by-Step Solution

   1. Multiply both sides by $x^3-1=(x-1)(x^2+x+1)$: $2x^2+1 = A(x^2+x+1) + (Bx+C)(x-1)$.
   2. Set $x=1$: $2(1)+1 = A(1+1+1) + 0 \Rightarrow 3 = 3A \Rightarrow A=1$.
   3. Expand the right side: $A x^2+Ax+A + Bx^2-Bx+Cx-C = (A+B)x^2+(A-B+C)x+(A-C)$.
   4. Match $x^2$ coefficient: $A+B=2 \Rightarrow B = 2-1=1$.
   5. Match constant term: $A - C = 1 \Rightarrow C = A-1 = 0$. …

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10. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
    Q.If $\dfrac{x+1}{(x-1)^2(x^2+1)} = \dfrac{A}{x-1} + \dfrac{B}{(x-1)^2} + \dfrac{Cx+D}{x^2+1}$, then $\sqrt{3A^2+4D^2+5C^2+B^2} =$ (A) $\dfrac{3}{2}$ (B) $\dfrac{1}{2}$ (C) 1 (D) 2

    ›Reveal solutionSolution

    Solving the partial-fraction decomposition gives $A=-\tfrac12, B=1, C=\tfrac12, D=-\tfrac12$; substituting into $\sqrt{3A^2+4D^2+5C^2+B^2}$ gives $2$.

    Concept and Intuition

    A rational function with a repeated linear factor $(x-1)^2$ and an irreducible quadratic factor $(x^2+1)$ decomposes as $\dfrac A{x-1}+\dfrac B{(x-1)^2}+\dfrac{Cx+D}{x^2+1}$. Clearing denominators and matching coefficients (or plugging convenient values of $x$) pins down all four constants.

    Step-by-Step Solution

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

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

    2. Plug $x=1$: LHS $=2$. RHS $= 0 + B(1+1) + 0 = 2B$. So $B=1$.
    3. Expand each term:
       • $A(x-1)(x^2+1) = A(x^3-x^2+x-1)$.
       • $B(x^2+1) = x^2+1$ (using $B=1$).
       • $(Cx+D)(x-1)^2 = (Cx+D)(x^2-2x+1) = Cx^3+(-2C+D)x^2+(C-2D)x+D$.
    4. Collect coefficients and match with $x+1 = 0\cdot x^3+0\cdot x^2+1\cdot x+1$:
       • $x^3$: $A+C=0$.
       • $x^2$: $-A+1-2C+D=0$.
       • $x^1$: $A+C-2D=1$.
       • $x^0$: $-A+1+D=1$.
    5. From $x^3$: $C=-A$. Substitute into $x^1$ equation: $A-A-2D=1 \Rightarrow D=-\tfrac12$.
    6. From $x^0$: $-A+D=0 \Rightarrow A=D=-\tfrac12$, hence $C=-A=\tfrac12$.
    7. (Verify $x^2$ equation: $-(-\tfrac12)+1-2(\tfrac12)+(-\tfrac12) = \tfrac12+1-1-\tfrac12=0$ ✓.) …

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11. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
    Q.If $\displaystyle\int \dfrac{x^3 - 1}{x^3 + x^2}\, dx = f(x) + \log(g(x)) + c$, $f(1) = 2$ and $g(-3) = \dfrac{3}{4}$, then $f(-2) + g(-2) =$ (A) $-\dfrac{9}{2}$ (B) $-\dfrac{1}{2}$ (C) $\dfrac{9}{4}$ (D) $\dfrac{1}{4}$

    ›Reveal solutionSolution

    A rational-function integral splits by partial fractions into a polynomial/rational part $f(x)$ plus a logarithmic part $\log(g(x))$; the two given data points pin down $f$ and $g$ exactly, letting us evaluate $f(-2)+g(-2)$. Answer: $-\tfrac12$.

    Concept and Intuition

    When an improper rational integrand is written $\frac{x^3-1}{x^3+x^2}$, polynomial division peels off the constant part, and partial fractions turn the remaining proper fraction into simple terms of the form $\frac{A}{x}, \frac{B}{x^2}, \frac{C}{x+1}$ whose antiderivatives are $A\log|x|$, $-B/x$, $C\log|x+1|$. Collecting all log terms into a single $\log(g(x))$ and all algebraic terms into $f(x)$ matches the form the question gives; the two numeric conditions are just there to confirm the constants (and resolve the sign inside the absolute value at negative $x$).

    Step-by-Step Solution

    1. Divide: $\dfrac{x^3-1}{x^3+x^2}=1-\dfrac{x^2+1}{x^3+x^2}=1-\dfrac{x^2+1}{x^2(x+1)}$.
    2. Partial fractions: $\dfrac{x^2+1}{x^2(x+1)}=\dfrac{A}{x}+\dfrac{B}{x^2}+\dfrac{C}{x+1}$. Clearing denominators: $x^2+1=Ax(x+1)+B(x+1)+Cx^2$. Setting $x=0$: $B=1$. Setting $x=-1$: $2=C$. Matching $x^2$ coefficients: $1=A+C\Rightarrow A=-1$.
    3. So $\dfrac{x^2+1}{x^2(x+1)}=-\dfrac1x+\dfrac1{x^2}+\dfrac2{x+1}$, and the integrand is $1+\dfrac1x-\dfrac1{x^2}-\dfrac2{x+1}$.
    4. Integrate term by term: $\displaystyle\int\left(1+\frac1x-\frac1{x^2}-\frac{2}{x+1}\right)dx=x+\log|x|+\frac1x-2\log|x+1|+c=\left(x+\frac1x\right)+\log\left|\frac{x}{(x+1)^2}\right|+c$. …

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12. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
    Q.If $\dfrac{A}{x-a} + \dfrac{Bx+C}{x^2+b^2} = \dfrac{1}{(x-a)(x^2+b^2)}$ then $C =$ (A) $\dfrac{-1}{a^2+b^2}$ (B) $\dfrac{1}{a^2+b^2}$ (C) $\dfrac{-a}{a^2+b^2}$ (D) $\dfrac{a}{a^2+b^2}$

    ›Reveal solutionSolution

    A standard partial-fractions comparison; clearing denominators and matching coefficients gives $C=\dfrac{-a}{a^2+b^2}$.

    Concept and Intuition

    When a rational function is split into partial fractions, multiplying through by the common denominator turns the identity into a polynomial identity, which must hold for all $x$. That lets us either substitute convenient values of $x$ or match coefficients of like powers of $x$.

    Step-by-Step Solution

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

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

    2. Put $x=a$ (kills the second term): $A(a^2+b^2)=1\Rightarrow A=\dfrac{1}{a^2+b^2}$.
    3. Expand the right side: $Ax^2+Ab^2+Bx^2-aBx+Cx-aC=1$.
    4. Coefficient of $x^2$: $A+B=0\Rightarrow B=-A=\dfrac{-1}{a^2+b^2}$. …

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