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Mathematics · 2026 · 3 marks
CBSE 2026 · Region 1 · Set 1 · Q27
(a) Find ∫ √((x+2)/(x-2)) d x
(a)
Find $\displaystyle \int \sqrt{\frac{x+2}{x-2}} \mathrm{~d} x$
OR
(b)
Find : $\displaystyle \int \frac{x^{2}}{\left(x^{2}+9\right)\left(x^{2}+16\right)} \mathrm{d} x$
Official answer
(a)
√(x² - $\displaystyle 4$) - 2sec⁻¹(|x|/$\displaystyle 2$) + C OR (b) ($\displaystyle 4$/$\displaystyle 7$)tan⁻¹(x/$\displaystyle 4$) - ($\displaystyle 3$/$\displaystyle 7$)tan⁻¹(x/$\displaystyle 3$) + C
Marking-scheme solution
$\displaystyle I = \int\sqrt{\dfrac{x+2}{x-2}}\,dx = \int\sqrt{\dfrac{x+2}{x-2}}\times\sqrt{\dfrac{x+2}{x+2}}\,dx = \int\dfrac{x+2}{\sqrt{x^2-4}}\,dx$
$\displaystyle = \dfrac12\int\dfrac{2x}{\sqrt{x^2-4}}\,dx + 2\int\dfrac{1}{\sqrt{x^2-4}}\,dx$
$\displaystyle I_1 = \dfrac12\int\dfrac{2x}{\sqrt{x^2-4}}\,dx = \dfrac12\int\dfrac{dt}{\sqrt{t}} = \sqrt{t} = \sqrt{x^2-4} \quad [(x^2-4)=t \Rightarrow 2x\,dx = dt]$
$\displaystyle I_2 = 2\int\dfrac{1}{\sqrt{x^2-4}}\,dx = 2\log\left|x+\sqrt{x^2-4}\right|$
$\displaystyle I = I_1 + I_2 = \sqrt{x^2-4} + 2\log\left|x+\sqrt{x^2-4}\right| + C$
Put $\displaystyle x^2=t$: $\displaystyle \dfrac{x^2}{(x^2+9)(x^2+16)} = \dfrac{t}{(t+9)(t+16)} = \dfrac{A}{t+9}+\dfrac{B}{t+16}$,
giving $\displaystyle A=-9/7$, $\displaystyle B=16/7$
Given integral $\displaystyle = -\dfrac97\int\dfrac{1}{x^2+9}\,dx + \dfrac{16}{7}\int\dfrac{1}{x^2+16}\,dx$
$\displaystyle = -\dfrac37\tan^{-1}\dfrac{x}{3} + \dfrac47\tan^{-1}\dfrac{x}{4} + C$
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CBSE Class 12 Mathematics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.