Q.Evaluate the definite integral: $\int_{1}^{2} \frac{5x^2}{x^2+4x+3} \, dx$

Showing the 12 most recent of 63 on this concept.

1. 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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2. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
   Q.Given $\dfrac{3x-2}{(x+1)^2(x+3)} = \dfrac{A}{x+1} + \dfrac{B}{(x+1)^2} + \dfrac{C}{x+3}$ then $4A + 2B + 4C$ (A) $5$ (B) $-5$ (C) $-3$ (D) $3$

   ›Reveal solutionSolution

   Standard partial-fraction cover-up plus a coefficient match pins down $A,B,C$, giving $4A+2B+4C=-5$.

   Concept and Intuition

   For a repeated linear factor $(x+1)^2$ together with a simple factor $(x+3)$, the cover-up (Heaviside) method quickly gives $B$ and $C$ by substituting the roots that make each factor vanish. The remaining constant $A$ is then found by matching a coefficient (here, the $x^2$ coefficient, since the numerator on the left has no $x^2$ term).

   Step-by-Step Solution

   1. Clear denominators: $3x-2 = A(x+1)(x+3) + B(x+3) + C(x+1)^2$.
   2. Set $x=-1$ (kills the $A$ and $C$ terms): $3(-1)-2 = B(-1+3) \Rightarrow -5 = 2B \Rightarrow B=-\dfrac{5}{2}$.
   3. Set $x=-3$ (kills the $A$ and $B$ terms): $3(-3)-2 = C(-3+1)^2 \Rightarrow -11 = 4C \Rightarrow C=-\dfrac{11}{4}$.
   4. Match the coefficient of $x^2$ on both sides (LHS has none): $0 = A + C \Rightarrow A = \dfrac{11}{4}$. …

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3. 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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4. 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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5. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
   Q.If the equivalent partial fraction of $\frac{x^3}{(2x-1)(x+2)(x-3)}$ is of the form $A + \frac{B}{2x-1} + \frac{C}{x+2} + \frac{D}{x-3}$ then the value of $A+B+C =$ (A) $-8/25$ (B) $4/25$ (C) $-1/50$ (D) $1/2$

   ›Reveal solutionSolution

   This is an improper partial fraction (numerator degree = denominator degree), so there's a constant term $A$ found from the leading behaviour, and $B,C$ are found by the standard cover-up (root-substitution) method — giving $A+B+C=4/25$.

   Concept and Intuition

   When the degree of the numerator equals the degree of the denominator, ordinary partial fractions leave a nonzero polynomial part (here, just a constant $A$, since both degrees are 3 and the denominator's leading coefficient is 2 — so $A$ equals the ratio of leading coefficients, $1/2$). The remaining proper-fraction coefficients ($B$, $C$, $D$) are then found efficiently using the "cover-up" trick: multiply through by the denominator and substitute each root of a linear factor to instantly isolate that factor's coefficient.

   Step-by-Step Solution

   1. As $x\to\infty$, $\dfrac{x^3}{(2x-1)(x+2)(x-3)} \to \dfrac{x^3}{2x^3} = \dfrac12$, so the constant part is $A=\dfrac12$.
   2. Multiply both sides by $(2x-1)(x+2)(x-3)$: $x^3 = A(2x-1)(x+2)(x-3) + B(x+2)(x-3) + C(2x-1)(x-3) + D(2x-1)(x+2)$.
   3. Set $x=\tfrac12$ (kills the $A$, $C$, $D$ terms): $\left(\tfrac12\right)^3 = B\left(\tfrac12+2\right)\left(\tfrac12-3\right) = B(2.5)(-2.5)=-6.25B$. $\tfrac18 = -\tfrac{25}{4}B \Rightarrow B = -\tfrac{1}{50}$. …

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6. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.If $\dfrac{13x+43}{2x^2+17x+30}=\dfrac{A}{2x+5}+\dfrac{B}{x+6}$ then $A+B=$ (A) 8 (B) 18 (C) 3 (D) 5

   ›Reveal solutionSolution

   This is a partial fractions decomposition; solving for $A$ and $B$ via convenient substitutions gives $A+B=8$.

   Concept and Intuition

   Once the denominator is factored into its linear pieces, the standard partial-fractions trick is to clear denominators and then substitute the root of each linear factor in turn — this isolates one unknown constant at a time without solving simultaneous equations.

   Step-by-Step Solution

   1. Factor the denominator: $2x^2+17x+30=(2x+5)(x+6)$ (check: $2x\cdot x + 2x\cdot6+5\cdot x+5\cdot6=2x^2+12x+5x+30=2x^2+17x+30$ ✓).
   2. Write $13x+43 = A(x+6)+B(2x+5)$.
   3. Set $x=-6$: $13(-6)+43=-78+43=-35 = A(0)+B(2(-6)+5)=B(-7) \Rightarrow B=5$. …

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7. 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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8. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.If $\dfrac{-x^2+6x+1}{(x-1)^2(x^2+2)} = \dfrac{A}{x-1}+\dfrac{B}{(x-1)^2}+\dfrac{Cx-3}{x^2+2}$, then $A+B+C=$ (A) $7$ (B) $5$ (C) $3$ (D) $2$

   ›Reveal solutionSolution

   Clearing denominators and matching coefficients (using $x=1$ to isolate $B$ first) gives $A=0,\ B=2,\ C=0$, so $A+B+C=2$.

   Concept and Intuition

   This is a standard partial-fractions decomposition. The repeated linear factor $(x-1)^2$ lets us find $B$ instantly by substituting $x=1$ directly into the cleared equation (this kills every term except the one multiplying $B$). The remaining coefficients $A,C$ are then found by matching powers of $x$.

   Step-by-Step Solution

   1. Clear denominators: $-x^2+6x+1=A(x-1)(x^2+2)+B(x^2+2)+(Cx-3)(x-1)^2$.
   2. Set $x=1$: LHS $=-1+6+1=6$; RHS $=0+B(3)+0=3B$. So $B=2$.
   3. Expand the RHS fully with $B=2$: $A(x^3-x^2+2x-2)+2x^2+4+(Cx-3)(x^2-2x+1)$. $(Cx-3)(x^2-2x+1)=Cx^3-2Cx^2+Cx-3x^2+6x-3$.
   4. Collect by power of $x$: $x^3$: $A+C$ $x^2$: $-A+2-2C-3=-A-2C-1$ $x^1$: $2A+C+6$ $x^0$: $-2A+4-3=-2A+1$
   5. Match to LHS coefficients ($0,-1,6,1$ for $x^3,x^2,x,1$): $x^3$: $A+C=0$ …

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9. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
   Q.The partial fraction of $\dfrac{x^2}{x^2 + 3x - 4}$ is ________ (A) $1 + \dfrac{-16}{5(x+4)} + \dfrac{1}{5(x-1)}$ (B) $1 + \dfrac{-1}{x+4} + \dfrac{1}{x-1}$ (C) $1 + \dfrac{-13}{5(x+4)} + \dfrac{1}{5(x-1)}$ (D) $\dfrac{2}{x+4} + \dfrac{1}{x-1}$

   ›Reveal solutionSolution

   Since the numerator's degree equals the denominator's degree, perform polynomial division first, then resolve the remaining proper fraction into partial fractions. Answer: (A).

   Concept and Intuition

   Partial fraction decomposition applies to a proper rational function (numerator degree less than denominator degree). Here both are degree 2, so we must first extract the constant (integer) part via division, leaving a proper fraction to decompose.

   Step-by-Step Solution

   1. Factor the denominator: $x^2+3x-4=(x+4)(x-1)$.
   2. Divide: $x^2 = (x^2+3x-4) - (3x-4)$, so $\dfrac{x^2}{x^2+3x-4} = 1 - \dfrac{3x-4}{(x+4)(x-1)}$.
   3. Decompose $\dfrac{3x-4}{(x+4)(x-1)} = \dfrac{A}{x+4}+\dfrac{B}{x-1}$, so $3x-4=A(x-1)+B(x+4)$.
   4. Set $x=1$: $-1 = 5B \Rightarrow B=-\dfrac15$.
   5. Set $x=-4$: $-16 = -5A \Rightarrow A=\dfrac{16}{5}$.
   6. So $\dfrac{3x-4}{(x+4)(x-1)} = \dfrac{16}{5(x+4)} - \dfrac{1}{5(x-1)}$. …

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10. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
    Q.If $\dfrac{13x+43}{2x^2+17x+30} = \dfrac{A}{2x+5} + \dfrac{B}{x+6}$, then $A^2+B^2 =$ (A) $22/3$ (B) $52$ (C) $34$ (D) $18/5$

    ›Reveal solutionSolution

    Factoring the denominator and matching coefficients gives $A=3$, $B=5$, so $A^2+B^2=34$.

    Concept and Intuition

    Partial fraction decomposition of a rational function with a factorable denominator reduces to writing the numerator as a linear combination of the "cleared" denominators of each partial fraction, then matching coefficients of like powers of $x$ (or substituting convenient values of $x$).

    Step-by-Step Solution

    1. Factor $2x^2+17x+30$: looking for factors of the form $(2x+5)(x+6)$, expand to check: $(2x+5)(x+6) = 2x^2+12x+5x+30 = 2x^2+17x+30$. ✓.
    2. Write $\dfrac{13x+43}{(2x+5)(x+6)} = \dfrac{A}{2x+5}+\dfrac{B}{x+6}$. Clearing denominators: $13x+43 = A(x+6) + B(2x+5)$.
    3. Expand the right side: $Ax+6A+2Bx+5B = (A+2B)x + (6A+5B)$.
    4. Match coefficients: $A+2B=13$ (coefficient of $x$) and $6A+5B=43$ (constant term). …

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11. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
    Q.If $\dfrac{17x-2}{12x^2-x-20} = \dfrac{A}{ax+5} + \dfrac{B}{3x+b}$ then $a.A + b.B =$ (A) 0 (B) 4 (C) 7 (D) 10

    ›Reveal solutionSolution

    Factoring the denominator and doing a partial-fraction decomposition gives $a=4,A=3,b=-4,B=2$, so $aA+bB=12-8=4$.

    Concept and Intuition

    Partial fraction decomposition requires first factoring the denominator to match the given linear-factor forms, then solving for the numerator constants by substituting the roots of each factor (the Heaviside cover-up method).

    Step-by-Step Solution

    1. Factor $12x^2-x-20$: looking for factors of $(4x+5)(3x-4)=12x^2-16x+15x-20=12x^2-x-20$. ✓
    2. Matching to the given form $\dfrac{A}{ax+5}+\dfrac{B}{3x+b}$: since one factor is $(4x+5)$, we get $a=4$; the other factor is $(3x-4)$, matching $3x+b$ gives $b=-4$.
    3. So $\dfrac{17x-2}{(4x+5)(3x-4)}=\dfrac{A}{4x+5}+\dfrac{B}{3x-4}$, i.e. $17x-2=A(3x-4)+B(4x+5)$.
    4. Set $x=\dfrac43$ (root of $3x-4=0$): $17\cdot\dfrac43-2=B\left(4\cdot\dfrac43+5\right)\Rightarrow\dfrac{68}{3}-\dfrac63=B\cdot\dfrac{31}{3}\Rightarrow\dfrac{62}{3}=\dfrac{31B}{3}\Rightarrow B=2$. …

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12. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
    Q.If $\dfrac{ax+5}{(x^2+b)(x+3)} = \dfrac{x+21}{12(x^2+b)} + \dfrac{c}{12(x+3)}$, then $b^2 =$ (A) $a^3-c$ (B) $a^2+c$ (C) $a-c$ (D) $a+c$

    ›Reveal solutionSolution

    Clear denominators in the partial-fraction identity, match coefficients of powers of $x$ to solve for $a$, $b$, $c$, then check which combination of $a,c$ equals $b^2$. Answer: $b^2=a^3-c$.

    Concept and Intuition

    A partial-fraction identity must hold for all $x$, so after clearing denominators the two sides are polynomials that must agree coefficient-by-coefficient. This converts a rational-function identity into a simple linear system.

    Step-by-Step Solution

    1. Multiply both sides by $12(x^2+b)(x+3)$:

    $12(ax+5) = (x+21)(x+3) + c(x^2+b)$

    2. Expand the right side: $(x+21)(x+3)=x^2+24x+63$, so RHS $=x^2+24x+63+cx^2+cb=(1+c)x^2+24x+(63+cb)$.
    3. LHS $=12ax+60$ has no $x^2$ term, so matching $x^2$ coefficients: $1+c=0\Rightarrow c=-1$.
    4. Matching $x$ coefficients: $12a=24\Rightarrow a=2$.
    5. Matching constants: $60=63+cb=63+(-1)b=63-b\Rightarrow b=3$. …

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