Aus Online Mathematik Brückenkurs 2

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

\displaystyle (x^2+2x+2)' = 2x+2 = 2(x+1)

so we can rewrite the integral as

\displaystyle \int \frac{\tfrac{1}{2}}{x^2+2x+2}\cdot (x^2+2x+2)'\,dx\,\textrm{.}

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{align} u &= x^2+2x+2\\[5pt] du &= (x^2+2x+2)'\,dx = 2(x+1)\,dx \end{align}\right\}\\[5pt] &= \frac{1}{2}\int \frac{du}{u}\\[5pt] &= \frac{1}{2}\ln |u| + C\\[5pt] &= \frac{1}{2}\ln |x^2+2x+2| + C\,\textrm{.} \end{align}

Note: By completing the square

\displaystyle x^2+2x+2 = (x+1)^2-1^2+2 = (x+1)^2+1

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

\displaystyle \frac{1}{2}\ln (x^2+2x+2) + C\,\textrm{.}