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Evaluate $\int \frac{x+1}{x^2+4 x+8} d x$.
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The quadratic denominator does not factor. We could complete the square and use a trigonometric substitution, but it is simpler to rearrange the integrand:
$\int \frac{x+1}{x^2+4 x+8} d x=\int \frac{x+2}{x^2+4 x+8} d x-\int \frac{1}{x^2+4 x+8} d x$
The first integral is an easy substitution problem, using $u=x^2+4 x+8$ :
$\int \frac{x+2}{x^2+4 x+8} d x=\frac{1}{2} \int \frac{d u}{u}=\frac{1}{2} \ln \left|x^2+4 x+8\right| .$
For the second integral we complete the square:
$x^2+4 x+8=(x+2)^2+4=4\left(\left(\frac{x+2}{2}\right)^2+1\right) \text {, }$
making the integral
$\frac{1}{4} \int \frac{1}{\left(\frac{x+2}{2}\right)^2+1} d x$
Using $u=\frac{x+2}{2}$ we get
$\frac{1}{4} \int \frac{1}{\left(\frac{x+2}{2}\right)^2+1} d x=\frac{1}{4} \int \frac{2}{u^2+1} d u=\frac{1}{2} \arctan \left(\frac{x+2}{2}\right) .$
The final answer is now
$\int \frac{x+1}{x^2+4 x+8} d x=\frac{1}{2} \ln \left|x^2+4 x+8\right|-\frac{1}{2} \arctan \left(\frac{x+2}{2}\right)+C .$
by Platinum (141,884 points)

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