CBSE 2024 · Region 3 · Set 1 · Q28 · 3 marks
Find : \[\int \frac{x^{2}+1}{\left(x^{2}+2\right)\left(x^{2}+4\right)} d x \]
Marking-scheme solution
$$I=\int \frac{x^{2}+1}{\left(x^{2}+2\right)\left(x^{2}+4\right)} d xLet $\displaystyle x^{2}=y$, then $\displaystyle \frac{x^{2}+1}{\left(x^{2}+2\right)\left(x^{2}+4\right)}=\frac{y+1}{(y+2)(y+4)}$
Let $\displaystyle \frac{y+1}{(y+2)(y+4)}=\frac{A}{y+2}+\frac{B}{y+4}$
this gives $\displaystyle A=-\frac{1}{2}, B=\frac{3}{2}$
$\displaystyle \therefore I=-\frac{1}{2} \int \frac{1}{x^{2}+2} d x+\frac{3}{2} \int \frac{1}{x^{2}+4} d x$
$\displaystyle \Rightarrow I=-\frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{x}{\sqrt{2}}\right)+\frac{3}{4} \tan ^{-1}\left(\frac{x}{2}\right)+c$
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CBSE Class 12 Mathematics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.