Aus Online Mathematik Brückenkurs 2

If we differentiate the denominator in the integrand

\displaystyle (x^2+1)' = 2x

we obtain almost the same expression as in the numerator; there is a constant 2 which is different. We therefore rewrite the numerator as

\displaystyle 3x = \frac{3}{2}\cdot 2x = \frac{3}{2}\cdot (x^2+1)',

so the integral can be written as

\displaystyle \int\frac{\tfrac{3}{2}}{x^2+1}\cdot (x^{2}+1)'\,dx\,,

and we see that the substitution \displaystyle u=x^2+1 can be used to simplify the integral,

\displaystyle \begin{align} \int \frac{3x}{x^2+1}\,dx &= \left\{ \begin{align} u &= x^2+1\\[5pt] du &= (x^2+1)'\,dx = 2x\,dx \end{align}\right\}\\[5pt] &= \frac{3}{2}\int \frac{du}{u}\\[5pt] &= \frac{3}{2}\ln |u|+C\\[5pt] &= \frac{3}{2}\ln |x^{2}+1|+C\\[5pt] &= \frac{3}{2}\ln (x^{2}+1) + C\,\textrm{.} \end{align}

In the last step, we take away the absolute sign around the argument in \displaystyle \ln, because \displaystyle x^2+1 is always greater than or equal to 1.