The value of \(\int \frac{1+x^{4}}{1+x^{6}} d x\) is

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 \(x^{3}=t\)
\(\begin{aligned}
\Rightarrow \quad 3 x^{2} d x &=d t \\
\Rightarrow \quad x^{2} d x &=\frac{d t}{3} \\
&=\tan ^{-1} x+\frac{1}{3} \tan ^{-1}(t)+C \\
&=\tan ^{-1} x+\frac{1}{3} \tan ^{-1}\left(x^{3}\right)+C
\end{aligned}\)