Aus Online Mathematik Brückenkurs 2

We could substitute \displaystyle u=x-\text{1}, but we would then still have the problem of the second term, \displaystyle \text{3} in the denominator. Instead, we take out a factor \displaystyle \text{3} from the denominator,

\displaystyle \begin{align} & \int{\frac{\,dx}{\left( x-1 \right)^{2}+3}}=\int{\frac{\,dx}{3\left( \frac{1}{3}\left( x-1 \right)^{2}+1 \right)}} \\ & =\frac{1}{3}\int{\frac{\,dx}{\frac{1}{3}\left( x-1 \right)^{2}+1}} \\ \end{align}

and move a factor \displaystyle \frac{1}{3} into the square \displaystyle \left( x-1 \right)^{2},

\displaystyle \frac{1}{3}\int{\frac{\,dx}{\frac{1}{3}\left( x-1 \right)^{2}+1}}=\frac{1}{3}\int{\frac{\,dx}{\left( \frac{x-1}{\sqrt{3}} \right)^{2}+1}}

Now, we substitute \displaystyle u=\frac{x-1}{\sqrt{3}} and get rid of all the problems at once:

\displaystyle \begin{align} & \frac{1}{3}\int{\frac{\,dx}{\left( \frac{x-1}{\sqrt{3}} \right)^{2}+1}}=\left\{ \begin{matrix} u=\frac{x-1}{\sqrt{3}} \\ du=\frac{\,dx}{\sqrt{3}} \\ \end{matrix} \right\} \\ & =\frac{1}{3}\int{\frac{\sqrt{3}\,du}{u^{2}+1}}=\frac{\sqrt{3}}{3}\int{\frac{\,du}{u^{2}+1}} \\ & =\frac{1}{\sqrt{3}}\arctan u+C \\ & =\frac{1}{\sqrt{3}}\arctan \frac{x-1}{\sqrt{3}}+C \\ \end{align}