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Mathematics · 2023 · 3 marks
CBSE 2023 · Region 1 · Set 3 · Q28
Find: ∫ (x^2)/(( x^2+4)( x^2+9)) d x
Find : \[\int \frac{\mathrm{x}^{2}}{\left(\mathrm{x}^{2}+4\right)\left(\mathrm{x}^{2}+9\right)} d \mathrm{x} \]
Marking-scheme solution
$$\begin{aligned}
& \frac{\mathrm{x}^{2}}{\left(\mathrm{x}^{2}+4\right)\left(\mathrm{x}^{2}+9\right)}=\frac{\mathrm{t}}{(\mathrm{t}+4)(\mathrm{t}+9)}=-\frac{4}{5} \frac{1}{\mathrm{t}+4}+\frac{9}{5} \frac{1}{\mathrm{t}+9} \\
&=-\frac{4}{5} \frac{1}{\left(\mathrm{x}^{2}+4\right)}+\frac{9}{5} \frac{1}{\left(\mathrm{x}^{2}+9\right)} \\
& \therefore \int \frac{\mathrm{x}^{2}}{\left(\mathrm{x}^{2}+4\right)\left(\mathrm{x}^{2}+9\right)} \mathrm{dx}=-\frac{4}{5} \int \frac{\mathrm{dx}}{\mathrm{x}^{2}+4}+\frac{9}{5} \int \frac{\mathrm{dx}}{\mathrm{x}^{2}+9} \\
&=-\frac{2}{5} \tan ^{-1} \frac{\mathrm{x}}{2}+\frac{3}{5} \tan ^{-1} \frac{\mathrm{x}}{3}+\mathrm{C}
\end{aligned}
$$
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CBSE Class 12 Mathematics past-paper question from the 2023 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.