Q52 / 120 · VITEEE 2011 · Mathematics
Evaluate $\int \frac{x^{2}+4}{x^{4}+16} \mathrm{dx}$.
- Correct answerA$\frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{x^{2}-4}{2 x \sqrt{2}}\right)+c$
- B$\frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{x^{2}-4}{2 \sqrt{2}}\right)+c$
- C$\frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{x^{2}-4}{x \sqrt{2}}\right)+c$
- DNone of the above
Solution
Let $\mathrm{I}=\int \frac{\mathrm{x}^{2}+4}{\mathrm{x}^{4}+16} \mathrm{dx}$ $=\int \frac{1+\frac{4}{\mathrm{x}^{2}}}{\mathrm{x}^{2}+\frac{16}{\mathrm{x}^{2}}} \mathrm{dx}=\int \frac{1+\frac{4}{\mathrm{x}^{2}}}{\left(\mathrm{x}-\frac{4}{\mathrm{x}}\right)^{2}+8} \mathrm{dx}$
Putting $x-\frac{4}{x}=t$
So that $\left(1+\frac{4}{\mathrm{x}^{2}}\right) \mathrm{dx}=\mathrm{dt}$
$$
\begin{array}{l}
\therefore \mathrm{I}=\int \frac{\mathrm{dt}}{\mathrm{t}^{2}+(2 \sqrt{2})^{2}} \\
=\frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{\mathrm{t}}{2 \sqrt{2}}\right)+\mathrm{C} \\
\Rightarrow \mathrm{I}=\frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{\mathrm{x}-\frac{4}{\mathrm{x}}}{2 \sqrt{2}}\right)+\mathrm{C} \\
=\frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{\mathrm{x}^{2}-4}{2 \mathrm{x} \sqrt{2}}\right)+\mathrm{C}
\end{array}
$$
Hence, the correct answer is option: 'A'.