Answer
$$
\int x\sec^2(x^2)dx
=\frac{1}{2}\tan x^2 +c
$$
Work Step by Step
Since $ u= x^2 $, then $ du=2x dx $ and hence, $$
\int x\sec^2(x^2)dx =\frac{1}{2}\int \sec^2 u d u= \frac{1}{2}\tan u +c\\
=\frac{1}{2}\tan x^2 +c
$$
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