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$\int \frac{x^2}{1+x^6} d x$ is equal to
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Verified Answer
The correct answer is:
$\frac{1}{3} \tan ^{-1}\left(x^3\right)+C$
Let $I=\int \frac{x^2}{1+\left(x^3\right)^2} d x$
Put $x^3=z \Rightarrow 3 x^2 d x=d z$
$$
I=\frac{1}{3} \int \frac{d z}{1+z^2}=\frac{1}{3} \tan ^{-1}(z)+C=\frac{1}{3} \tan ^{-1}\left(x^3\right)+C
$$
Put $x^3=z \Rightarrow 3 x^2 d x=d z$
$$
I=\frac{1}{3} \int \frac{d z}{1+z^2}=\frac{1}{3} \tan ^{-1}(z)+C=\frac{1}{3} \tan ^{-1}\left(x^3\right)+C
$$
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