Q112 / 180 · COMEDK 2018 · Mathematics
$\int \frac{x^{2}+1}{x^{4}+1} d x$
- A$\frac{1}{\sqrt{2}} \log _{e}\left(x^{2}+1\right)+c$
- B$\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x^{2}+1}{x \sqrt{2}}\right)+c$
- C$\frac{1}{\sqrt{2}} \tan ^{-1}\left(x^{2}-1\right)+c$
- Correct answerD$\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x^{2}-1}{x \sqrt{2}}\right)+c$
Solution
We have, $\int \frac{x^{2}+1}{x^{4}+1} d x$
Divide numerator and denominator by $x^{2}$, we get
$$
\begin{aligned}
\int \frac{1+\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}} d x &=\int \frac{1+\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}-2+2} d x \\
&=\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}+2} d x
\end{aligned}
$$
Let $x-\frac{1}{x}=t$
$$
\begin{aligned}
\left(1+\frac{1}{x^{2}}\right) d x=d t &=\int \frac{d x}{t^{2}+2} \\
&=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{t}{\sqrt{2}}\right)+c \\
&=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)+c \\
&=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x^{2}-1}{\sqrt{2} x}\right)+c
\end{aligned}
$$
Hence, the correct answer is option: 'D'.