Q68 / 180 · KCET 2020 · Mathematics
The value of $\int \frac{1+x^{4}}{1+x^{6}} d x$ is
- A$\tan ^{-1} x+\tan ^{-1} x^{3}+C$
- Correct answerB$\tan ^{-1} x+\frac{1}{3} \tan ^{-1} x^{3}+C$
- C$\tan ^{-1} x-\frac{1}{3} \tan ^{-1} x^{3}+C$
- D$\tan ^{-1} x+\frac{1}{3} \tan ^{-1} x^{2}+C$
Solution
We have, $\begin{aligned} \int \frac{1+x^{4}}{1+x^{6}} d x \\=& \int \frac{1+x^{4}}{1+x^{6}} \times \frac{1+x^{2}}{1+x^{2}} d x \\=& \int \frac{1+x^{2}+x^{4}+x^{6}}{\left(1+x^{6}\right)\left(1+x^{2}\right)} d x \\=& \int \frac{1+x^{6}}{\left(1+x^{6}\right) 1+x^{2}} d x+\int \frac{x^{2}\left(1+x^{2}\right)}{\left(1+x^{6}\right)\left(1+x^{2}\right)} d x \\=& \int \frac{d x}{1+x^{2}}+\int \frac{x^{2}}{1+x^{6}} d x \\=& \tan ^{-1} x+\int \frac{x^{2}}{1+\left(x^{3}\right)^{2}} d x \\=& \tan ^{-1} x+\int \frac{d t / 3}{1+t^{2}} \end{aligned}$
Put $x^{3}=t$
$\begin{aligned}
\Rightarrow \quad 3 x^{2} d x &=d t \\
\Rightarrow \quad x^{2} d x &=\frac{d t}{3} \\
&=\tan ^{-1} x+\frac{1}{3} \tan ^{-1}(t)+C \\
&=\tan ^{-1} x+\frac{1}{3} \tan ^{-1}\left(x^{3}\right)+C
\end{aligned}$
Hence, the correct answer is option: 'B'.