Aus Online Mathematik Brückenkurs 2

What makes our integral differ from that in the exercise´s text is that there is a term \displaystyle x^{2}+4 instead of \displaystyle x^{2}+1, but if we factor out the 4 from the denominator,

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

we obtain the correct second term in the denominator. On the other hand, there is no longer \displaystyle x^{2} but \displaystyle \frac{1}{4}x^{2}, although we can get around this by substituting \displaystyle u=\frac{1}{2}x,

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