Integration with substitution I want to integrate (2x+1)/((x^2 - 6x + 14)^3)
I'm guessing you use substitution but im unsure what to substitute, is it best to make u = x^2 or X^2-6x or even x^2 - 6x + 14
I find that it makes turning the top of the fraction into terms of u very difficult
 A: Using Trigonometric substitution 
put $x-3=\sqrt5\tan\theta$ as $\displaystyle x^2-6x+14=(x-3)^2+(\sqrt5)^2$
$$\int \frac{2x+1}{(x^2-6x+14)^3}dx=\int\frac{2(\sqrt5\tan\theta+3)+1}{(5\sec^2\theta)^3}\sqrt5\sec^2\theta d\theta$$
$$=\frac1{25\sqrt5}\int(2\sqrt5\tan\theta+7)\cos^4\theta d\theta=\cdots$$
$$=\frac2{25}\int\sin\theta\cos^3\theta d\theta+\frac7{25\sqrt5}\int\cos^4\theta d\theta$$
The first integral can be managed by substituting $\cos\theta =u$
For the second one use Double-Angle Formulas  , $\displaystyle \cos^4\theta=\frac{(2\cos^2\theta)^2}4=\frac{(1+\cos2\theta)^2}4=\frac14+\frac{\cos2\theta}2+\frac{\cos^22\theta}4$ 
Again, $\displaystyle2\cos^22\theta=1+\cos4\theta$
As the principal value $\arctan$ lies $\left[-\frac\pi2,\frac\pi2\right]$
$\displaystyle\cos\theta=+\frac1{\sqrt{1+\tan^2\theta}}=\frac{\sqrt5}{\sqrt{x^2-6x+14}}$
and $\displaystyle\sin\theta=\tan\theta\cdot\cos\theta$
A: $$\begin{align} \int \frac{2x + 1} {(x^2 - 6x + 14)^3} & = \int \frac{2x - 6 + 6 + 1}{(x^2 - 6x + 14)^3}\,dx \\ \\ & = \int\dfrac{2x - 6}{(x^2 - 6x + 14)^3}\,dx + \int \frac{7}{(x^2 - 6x + 14)^3}\,dx \\ \\ &= \int\dfrac{\overbrace{2x - 6}^{du}}{(\underbrace{x^2 - 6x + 14}_{u})^3}\,dx + \int \frac{7}{(\underbrace{x^2 - 6x + 9}_{(x - 3)^2} + \underbrace{5}_{(\sqrt 5)^2})^3}\,dx \end{align}$$
The left-most integral: using $u = x^2 - 6x + 14 \implies du = 2x - 6, \,dx$
so we integrate $\displaystyle \int \frac {du}{u^3}$.
The right-most integral, we can use the substitution $x - 3 = \sqrt 5\tan \theta \implies dx = \sqrt 5 \sec^2 \theta\,d\theta$, to obtain the integral $$\displaystyle 7\int \frac{\sqrt 5 \sec^2 \theta \,d\theta}{(5 \tan^2 \theta + 5)^3}= 7\int \frac{\sqrt 5 \sec^2 \theta \,d\theta}{(5 (\tan^2 \theta +1))^3} =  7\int \frac{\sqrt 5 \sec^2 \theta \,d\theta}{(5 \sec^2\theta)^3} = \cdots$$
A: I'd suggest using $y = x^2 - 6x + 14$. Then $du = 2x - 6$, which is almost what you've got in the numerator. So split your fraction: 
$$
\frac{2x + 1} {(x^2 - 6x + 14)^3} = \frac{2x - 6} {(x^2 - 6x + 14)^3} + \frac{7} {(x^2 - 6x + 14)^3}
$$
That splits your integral into two parts; the first part can be integrated with the substitution I suggested. The second part will require a bit more work, completing the square in the denominator and doing a tangent-substitution, probably. 
