## Calculus 8th Edition

$\displaystyle \frac{1}{2}\ln|x^2+2x+5|+3\tan^{-1}(\frac{x+2}{2})+C$
**Partial fraction decomposition is not needed for this problem. Only u-sub and inverse trigonometric substitution is needed** $\displaystyle \int\frac{x+4}{x^2+2x+5}dx$ Split this fraction into the following parts: $\displaystyle \int\frac{x+1}{x^2+2x+5}+\frac{3}{x^2+2x+5}dx$ $\displaystyle \int\frac{x+1}{x^2+2x+5}dx+\int\frac{3}{x^2+2x+5}dx$ The first integral can be solved using u-sub as follows: $\displaystyle \int\frac{x+1}{x^2+2x+5}dx$ where$\quad u=x^2+2x+5\quad \rightarrow \quad du=(2x+2)dx = 2(x+1)dx$ $\displaystyle \frac{1}{2}\int\frac{du}{u} = \frac{1}{2}\ln|u|+C=\frac{1}{2}\ln|x^2+2x+5|+C$ The second integral can be solved using inverse trigonometric substitution as follows: Recall that $\displaystyle \int\frac{du}{a^2+u^2}=\frac{1}{a}\tan^{-1}(\frac{u}{a})+C$ $\displaystyle \therefore\int\frac{3}{x^2+2x+5}dx=\int\frac{3}{(x^2+2x+1)+4}dx=\int\frac{3}{(x+1)^2+4}dx$ where $\quad a^2=4$ and $u^2=(x+1)^2\quad\rightarrow\quad u=2$ and $a=x+1$ $\displaystyle \int\frac{3}{(x+1)^2+4}dx = 3(\frac{1}{2}\tan^{-1}(\frac{x+1}{2}))+C$ Adding both solutions gives us $\displaystyle \frac{1}{2}\ln|x^2+2x+5|+\frac{3}{2}\tan^{-1}(\frac{x+1}{2})+C$