Aus Online Mathematik Brückenkurs 2

The trick is to complete the square in the denominator so that we obtain the same expression as in exercise b,

\displaystyle \int{\frac{\,dx}{x^{2}+4x+8}}=\int{\frac{\,dx}{\left( x+2 \right)^{2}-2^{2}+8}}=\int{\frac{\,dx}{\left( x+2 \right)^{2}+4}}

We take out a factor \displaystyle \text{4} from the denominator

\displaystyle \int{\frac{\,dx}{\left( x+2 \right)^{2}+4}}=\int{\frac{\,dx}{4\left( \frac{1}{4}\left( x+2 \right)^{2}+1 \right)}}=\frac{1}{4}\int{\frac{\,dx}{\frac{1}{4}\left( x+2 \right)^{2}+1}}

and rewrite the quadratic term as

\displaystyle \frac{1}{4}\int{\frac{\,dx}{\frac{1}{4}\left( x+2 \right)^{2}+1}}=\frac{1}{4}\int{\frac{\,dx}{\left( \frac{x+2}{2} \right)^{2}+1}}

If we now substitute \displaystyle u=\frac{x+2}{2}, we obtain the integral in the exercise

\displaystyle \begin{align} & \frac{1}{4}\int{\frac{\,dx}{\left( \frac{x+2}{2} \right)^{2}+1}}=\left\{ \begin{matrix} u=\frac{x+2}{2} \\ du=\frac{\,dx}{2} \\ \end{matrix} \right\} \\ & =\frac{1}{4}\int{\frac{\,2dx}{u^{2}+1}}=\frac{1}{2}\int{\frac{\,dx}{u^{2}+1}} \\ & =\frac{1}{2}\arctan u+C \\ & =\frac{1}{2}\arctan \frac{x+2}{2}+C \\ \end{align}