Answer
$$\int\frac{1+x}{1+x^2}dx=\tan^{-1}x+\frac{1}{2}\ln|1+x^2|+C$$
Work Step by Step
$$A=\int\frac{1+x}{1+x^2}dx$$
Sometimes, the numerator of the fraction inside the integral might contain several elements like this one. You might try to break down the elements into several integrals and see if it would be easier to transform to familiar forms or not. $$A=\int [\frac{1}{1+x^2}+\frac{x}{1+x^2}]dx$$ $$A=\int\frac{1}{1+x^2}dx+\int\frac{x}{1+x^2}dx$$ $$A=B+C$$
*Consider $B$: $$B=\int\frac{1}{1+x^2}dx$$ $$B=\tan^{-1}x+C$$
*Consider $C$: $$C=\int\frac{x}{1+x^2}dx$$
Let $u=1+x^2$
Then we have $du=2xdx$. That means $xdx=\frac{1}{2}du$
Substitute into $C$: $$C=\frac{1}{2}\int\frac{1}{u}du$$ $$C=\frac{1}{2}\ln|u|+C$$ $$C=\frac{1}{2}\ln|1+x^2|+C$$
Overall, $$A=\tan^{-1}x+\frac{1}{2}\ln|1+x^2|+C$$
