Answer
$$\int\frac{\sin2x}{1+\cos^2 x}dx=-\ln(1+\cos^2 x)+C$$
Work Step by Step
$$A=\int\frac{\sin2x}{1+\cos^2 x}dx$$
Let $u=1+\cos^2x $
Then apply the Chain Rule, we have $du=(1+\cos^2 x)'dx=2\cos x(\cos x)'dx=(-2\cos x\sin x)dx=-\sin2xdx$.
That means $\sin2xdx=-du$
Substitute into $A$: $$A=-\int\frac{1}{u}du$$ $$A=-\ln|u|+C$$ $$A=-\ln|1+\cos^2x|+C$$
For all $x\in R$, $\cos^2x\geq0$, so it follows that $(1+\cos^2 x)\gt0$.
Therefore, for all $x\in R$, $|1+\cos^2 x|=1+\cos^2 x$
So, $$A=-\ln(1+\cos^2 x)+C$$
