Aus Online Mathematik Brückenkurs 2

Observe that the derivative of the denominator is, for the most part, equal to the numerator,

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

so we can rewrite the integral as

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

The substitution \displaystyle u=x^{2}+2x+2 will therefore simplify the integral considerably:

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

NOTE: By completing the square

\displaystyle x^{2}+2x+2=\left( x+1 \right)^{2}-1^{2}+2=\left( x+1 \right)^{2}+1

we see that \displaystyle x^{2}+2x+2 is always greater than or equal to \displaystyle \text{1}, so we can take away the absolute sign around the argument in \displaystyle \text{ln} and answer with

\displaystyle \frac{1}{2}\ln \left( x^{2}+2x+2 \right)+C