$\int \frac{2 x^2-1}{x^4-x^2-20} d x=$
$\int \frac{2 x^2-1}{x^4-x^2-20} d x=$
- $\frac{1}{\sqrt{5}} \log \left|\frac{\mathrm{x}+\sqrt{5}}{\mathrm{x}-\sqrt{5}}\right|+\tan ^{-1}\left(\frac{\mathrm{x}}{2}\right)+\mathrm{c}$
- $\frac{1}{2 \sqrt{5}} \log \left|\frac{\mathrm{x}+\sqrt{5}}{\mathrm{x}-\sqrt{5}}\right|+\tan ^{-1}\left(\frac{\mathrm{x}}{2}\right)+\mathrm{c}$
- $\frac{1}{2 \sqrt{5}} \log \left|\frac{\mathrm{x}-\sqrt{5}}{\mathrm{x}+\sqrt{5}}\right|+\frac{1}{2} \tan ^{-1}\left(\frac{\mathrm{x}}{2}\right)+\mathrm{c}$
- $\frac{1}{2} \log \left|\frac{\mathrm{x}-\sqrt{5}}{\mathrm{x}+\sqrt{5}}\right|+\frac{1}{2} \tan ^{-1}\left(\frac{\mathrm{x}}{2}\right)+\mathrm{c}$
Solution
Let $I=\int \frac{2 x^2-1}{x^4-x^2-20} d x$
If $x^2=t$, then $\frac{2 x^2-1}{x^4-x^2-20}=\frac{2 t-1}{t^2-t-20}$
Let $\frac{2 t-1}{(t-5)(t+4)}=\frac{A}{(t-5)}+\frac{B}{(t+4)}$
$\begin{aligned}
& \therefore 2 \mathrm{t}-1=(\mathrm{t}+4) \mathrm{a}+(\mathrm{t}-5) \mathrm{B} \\
& \therefore 2=\mathrm{A}+\mathrm{B} \text { and }-1=4 \mathrm{~A}-5 \mathrm{~B}
\end{aligned}$
Solving, we get $\mathrm{B}=1, \mathrm{~A}=1$
$\therefore I=\int\left[\frac{1}{x^2-5}+\frac{1}{x^2+4}\right] d x=\frac{1}{2 \sqrt{5}} \log \left|\frac{x-\sqrt{5}}{x+\sqrt{5}}\right|+\frac{1}{2} \tan ^{-1}\left(\frac{x}{2}\right)+c$
Asked in: MHT CET 2021 (20 Sep Shift 2)
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