Answer
$$\int x\sec x^2dx=\frac{1}{2}\ln|\sec x^2+\tan x^2|+C$$
Work Step by Step
$$A=\int x\sec x^2dx$$
We set $a=x^2$, which means $$da=2xdx$$ $$xdx=\frac{1}{2}da$$
Therefore, $$A=\frac{1}{2}\int \sec ada$$ $$A=\frac{1}{2}\ln|\sec a+\tan a|+C$$ $$A=\frac{1}{2}\ln|\sec x^2+\tan x^2|+C$$
