Aus Online Mathematik Brückenkurs 2

If we differentiate the denominator in the integrand

\displaystyle 3x=\frac{3}{2}\centerdot 2x=\frac{3}{2}\left( x^{2}+1 \right)^{\prime }

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

\displaystyle 3x=\frac{3}{2}\centerdot 2x=\frac{3}{2}\centerdot \left( x^{2}+1 \right)^{\prime },

so the integral can be written as

\displaystyle \int{\frac{\frac{3}{2}}{x^{2}+1}}\centerdot \left( x^{2}+1 \right)^{\prime }\,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{matrix} u=x^{2}+1 \\ du=\left( x^{2}+1 \right)^{\prime }\,dx=2x\,dx \\ \end{matrix} \right\} \\ & =\frac{3}{2}\int{\frac{\,du}{u}}=\frac{3}{2}\ln \left| u \right|+C \\ & =\frac{3}{2}\ln \left| x^{2}+1 \right|+C \\ & =\frac{3}{2}\ln \left( x^{2}+1 \right)+C \\ \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 \displaystyle \text{1}.