Integrate $\int \frac{6x-15}{x^2 +8x+101}dx$ I've been trying this one for days and I can't seem to get it. Any ideas?
$\int \frac{6x-15}{x^2 +8x+101}dx$
Thanks,
Nick
 A: Hint: $$\Large \frac{6x-15}{x^2+8x+101} \equiv 3\left[\frac{2x+8}{x^2+8x+101}\right]-39 \left[\frac{1}{x^2+8x+101}\right].$$
The first part should be easy to integrate (inspection): the second requires a trigonometric substitution.
Hint 2: $$\frac{1}{x^2+8x+101} \equiv \frac{1}{\color{green}{(x+4)^2}+\color{red}{85}}.$$
Now let $x+4 \equiv \sqrt{85} \tan(\theta).$

If Hint 2 is too tedious (i.e. you can't be bothered going through the whole process of substitution), you might want to use the standard result $$\int \frac{1}{\color{green}{u^2}+\color{red}{a^2}}\ \mathrm{d}u =\frac{1}{a}\arctan\left(\frac{u}{a}\right)+C \quad.$$
A: Hint; Write the integral as $$\int \frac{3(2x+8)}{x^2+8x+101}-\frac{39}{x^2+8x+101} dx$$
For the first let $$u=x^2+8x+101$$ and for the second rewrite it as $$39 \int \frac{1}{(x+4)^2+85} dx$$ and let $$v=x+4$$
A: $\displaystyle\int\dfrac{6x-15}{x^2+8x+101} dx=\int\dfrac{6x-15}{(x+4)^2+85} dx$.
Now let $u=x+4$, so $x=u-4$ and $dx=du$ to get 
$\displaystyle\int\frac{6u-39}{u^2+85} du=3\int\frac{2u}{u^2+85} du-39\int\frac{1}{u^2+85} du$
$=3\ln(u^2+85)-39\left(\frac{1}{\sqrt{85}}\arctan\frac{u}{\sqrt{85}}\right)+C$.
(Now substitute back for u.)
