# Calc1.50

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$\int {\frac {2x}{x^{2}+1}}\,dx\,$

To find this integral, start by making a substitution.

Let $u=x^{2}+1\,$

Then $du=2xdx\,$

Now solve for $dx\,$ to get

$dx={\frac {du}{2x}}\,$

Now substitute these functions into the integral.

$\int {\frac {2x}{x^{2}+1}}\,dx=\int {\frac {2x}{u}}{\frac {du}{2x}}=\int {\frac {du}{u}}\,$

Now, this integral is of a form of which we already know the technique.

$\int {\frac {du}{u}}=\ln |u|+C\,$

The final step is to undo the substition. Plug $x^{2}+1\,$ back in wherever there is a $u\,$. Also, note the absolute value bars can be dropped because $x^{2}+1\,$ is always positive. So our answer is

$\ln(x^{2}+1)+C\,$