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Q. The value of $\int\frac{1+x^{4}}{1+x^{6}}dx$ is

KCETKCET 2020Integrals

Solution:

$ \int \frac{1+x^{4}}{1+x^{6}} d x =\int \frac{1+x^{4}-x^{2}+x^{2}}{\left(1+x^{2}\right)\left(1-x^{2}+x^{4}\right)} d x$
$=\int\left(\frac{1-x^{2}+x^{4}}{\left(1+x^{2}\right)\left(1-x^{2}+x^{4}\right)}+\frac{x^{2}}{\left(1+x^{2}\right)\left(1-x^{2}+x^{4}\right)}\right) d x $
$=\int\left(\frac{1}{1+x^{2}}+\frac{1}{3} \cdot \frac{3 x^{2}}{1+\left(x^{3}\right)^{2}}\right) d x $
$=\tan ^{-1} x+\frac{1}{3} \tan ^{-1} x^{3}+C $