Evaluate: $\displaystyle\int \dfrac{2x^2-3}{(x^2-5)(x^2+4)}\,dx$

Evaluate: ∫ (2x²-3)/((x²-5)(x²+4))dx

Question 243 of 255

\displaystyle\int \dfrac{2x^2-3}{(x^2-5)(x^2+4)}\,dx

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2023Subjective· 4mImportance★★★★★

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Put $t=x^2$ , split into partial fractions, integrate each standard form.

Let $t=x^2$ : $\dfrac{2t-3}{(t-5)(t+4)}=\dfrac{A}{t-5}+\dfrac{B}{t+4}$

$2t-3=A(t+4)+B(t-5)$ . At $t=5$ : $7=9A\Rightarrow A=\dfrac79$ . At $t=-4$ : $-11=-9B\Rightarrow B=\dfrac{11}9$

So integrand $=\dfrac79\cdot\dfrac1{x^2-5}+\dfrac{11}9\cdot\dfrac1{x^2+4}$

$\displaystyle\int\dfrac{dx}{x^2-5}=\dfrac1{2\sqrt5}\log\left|\dfrac{x-\sqrt5}{x+\sqrt5}\right|+c_1$

$\displaystyle\int\dfrac{dx}{x^2+4}=\dfrac12\tan^{-1}\dfrac x2+c_2$

…

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 mark
$\int \frac{x}{(1 + x)(2 + x)}\,dx = \int \left(\frac{A}{1 + x} + \frac{B}{2 + x}\right)dx$
›Reveal solutionSolution

Partial fractions give $A = -1,\; B = 2$ , so the claim $A = 1,\, B = 2$ is False.

Decompose the integrand:

$\frac{x}{(1 + x)(2 + x)} = \frac{A}{1 + x} + \frac{B}{2 + x}.$

Multiply through by $(1+x)(2+x)$ :

$x = A(2 + x) + B(1 + x).$

Put $x = -1$ : $\;-1 = A(2 - 1) + B(0) = A \Rightarrow A = -1.$

…

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CBSE 2024Set ANNUAL1 markMCQ
$\displaystyle\int \dfrac{dx}{(x - 8)(x + 7)} =$
(a) $\dfrac{1}{15} \log \left|\dfrac{x + 2}{x - 1}\right| + c$
(b) $\dfrac{1}{15} \log \left|\dfrac{x + 8}{x + 7}\right| + c$
(c) $\dfrac{1}{15} \log \left|\dfrac{x - 8}{x + 7}\right| + c$
(d) $(x - 8)(x - 7) + c$
(e) $\dfrac{1}{15} \log \left|\dfrac{x + 2}{x + 1}\right| + c$
(f) $(x - 8)(x + 7) + c$
›Reveal solutionSolution

Partial fractions give $\dfrac{1}{(x-8)(x+7)} = \dfrac{1}{15}\!\left(\dfrac{1}{x-8} - \dfrac{1}{x+7}\right)$ , which integrates to $\dfrac{1}{15}\log\left|\dfrac{x-8}{x+7}\right| + c$ .

Write

$\frac{1}{(x-8)(x+7)} = \frac{A}{x-8} + \frac{B}{x+7} \;\Rightarrow\; A(x+7) + B(x-8) = 1.$

Put $x = 8$ : $\;A(15) = 1 \Rightarrow A = \dfrac{1}{15}$ .

Put $x = -7$ : $\;B(-15) = 1 \Rightarrow B = -\dfrac{1}{15}$ .

Hence …

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CBSE 2022Set ANNUAL1 mark
$\int \dfrac{x}{(x + 2)(x + 3)}\, dx =$
›Reveal solutionSolution
Partial fractions give $\dfrac{x}{(x+2)(x+3)} = \dfrac{-2}{x+2} + \dfrac{3}{x+3}$ , so the blank preceding $\displaystyle\int \dfrac{3}{x+3}\,dx$ is $\displaystyle\int \dfrac{-2}{x+2}\,dx$ .

Resolve the integrand into partial fractions:

$\frac{x}{(x+2)(x+3)} = \frac{A}{x+2} + \frac{B}{x+3}.$

Clearing denominators: $x = A(x+3) + B(x+2)$ .

Put $x = -2$ : $-2 = A(1) \Rightarrow A = -2$ .

Put $x = -3$ : $-3 = B(-1) \Rightarrow B = 3$ .

Hence

$\frac{x}{(x+2)(x+3)} = \frac{-2}{x+2} + \frac{3}{x+3},$

so …
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Q.Evaluate: $\displaystyle\int \dfrac{2x^2-3}{(x^2-5)(x^2+4)}\,dx$

Concept understanding — Integration by Partial Fractions

Q.Evaluate: ∫2x2−3(x2−5)(x2+4) dx\displaystyle\int \dfrac{2x^2-3}{(x^2-5)(x^2+4)}\,dx∫(x2−5)(x2+4)2x2−3​dx

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Q.Evaluate: $\displaystyle\int \dfrac{2x^2-3}{(x^2-5)(x^2+4)}\,dx$

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