I know the formula$$\int\frac{dx}{x^2+1} = \tan^{-1}(x) + C$$
But, when integrating by parts:
$$u = x^2+1$$ $$u' = 2x$$
$$v' = 1$$ $$v = x$$
$$\int\frac{dx}{x^2+1} = uv - \int vu' dx = x(x^2+1) - \int 2x^2dx$$ $$ \int 2x^2dx = \frac{2x^3}{3} + C$$
So, finally: $$\int\frac{dx}{x^2+1} = x(x^2+1) - \frac{2x^3}{3} + C$$
Am I doing something wrong?
Integration by parts tells us that:
$$ \int u v' \, dx = uv - \int vu' \, dx $$
What you did was: $$ \int \frac{v'}{u} \, dx \neq uv - \int vu' \, dx $$
Yes you are:
$$ u = x^2+1 $$ This seems a little fishy, notice that this part is in fraction?!? If you want to use partial integration you should have used:
$$ u = \frac{1}{x^2+1} $$
Look this is an easier approach: Put $x = \tan \alpha $ . then $dx = \sec^2 \alpha d \alpha $. Hence
$$ \int \frac{dx}{1 + x^2} = \int \frac{\sec^2 \alpha}{1 + \tan^2 \alpha} d \alpha = \int d \alpha = \alpha = \arctan x + C$$