Answer
$\frac{1}{2}\tan (x^2)+C$
Work Step by Step
Let $u=x^2$. Then $du=2xdx$, so $xdx=\frac{du}{2}$.
$\int x\sec^2 (x^2) dx$
$=\int \sec^2(x^2)xdx$
$=\int \sec^2 u \frac{du}{2}$
$=\frac{1}{2}\tan u+C$
$=\frac{1}{2}\tan (x^2)+C$
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