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Mathematics · 2022 · 4 marks
CBSE 2022 · Region 3 · Set 1 · Q12
Find: ∫ (x^2)/((x^2+1)(3 x^2+4)) d x OR Evaluate: ∫_-2^1 √(5-4 x-x^2) d x
Find: $\displaystyle \int \frac{x^{2}}{\left(x^{2}+1\right)\left(3 x^{2}+4\right)} \mathrm{d} x$
OR Evaluate: $\displaystyle \int_{-2}^{1} \sqrt{5-4 x-x^{2}} \mathrm{~d} x$
Marking-scheme solution
\[\int \frac{x^{2}}{(x^{2}+1)(3x^{2}+4)}\,dx=\int -\frac{1}{x^{2}+1}\,dx+\int \frac{4}{3x^{2}+4}\,dx
\]\[= -\tan^{-1}x+\frac{4}{3}\times\frac{\sqrt{3}}{2}\tan^{-1}\frac{\sqrt{3}x}{2}+C
\]\[= -\tan^{-1}x+\frac{2}{\sqrt{3}}\tan^{-1}\frac{\sqrt{3}x}{2}+C
\]OR(b)\[\int_{-2}^{1}\sqrt{5-4x-x^{2}}\,dx
\]\[=\int_{-2}^{1}\sqrt{9-(x+2)^{2}}\,dx
\]\[=\left[\frac{x+2}{2}\sqrt{9-(x+2)^{2}}+\frac{9}{2}\sin^{-1}\frac{x+2}{3}\right]_{-2}^{1}
\]\[= 0+\frac{9}{2}\cdot\frac{\pi}{2}-0
\]\[=\frac{9\pi}{4}
\]
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CBSE Class 12 Mathematics past-paper question from the 2022 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.