# What is the solution for : integrate ( 1/(9-12*x+4*x^2)) ?

May 25, 2017

$- \frac{1}{4 x + 6} + C$

#### Explanation:

Let's write the question formatted:

$\int \frac{1}{9 - 12 x + 4 {x}^{2}} \mathrm{dx}$

First off, we can simplify the denominator

$\frac{1}{9 - 12 x + 4 {x}^{2}} = \frac{1}{2 x + 3} ^ 2$

First, we are going to integrate by parts (u-sub)

$\int f \left(u\right) \frac{\mathrm{du}}{\mathrm{dx}} \mathrm{dx} = \int f \left(u\right) \mathrm{du}$

Let $u = 2 x + 3 , \setminus \quad \frac{\mathrm{du}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left[2 x + 3\right] = 2$
We get:

$\frac{1}{2} \int \frac{1}{u} ^ 2 \mathrm{du} = \frac{1}{2} \int {u}^{- 2} \mathrm{du}$

Solve.

$\frac{1}{2} \int {u}^{- 2} \mathrm{du} = - \frac{1}{2 u} + C$

Sub $u$ back.

$- \frac{1}{2 u} + C = - \frac{1}{2 \left(2 x + 3\right)} + C = - \frac{1}{4 x + 6} + C$

$\therefore$ $\int \frac{1}{9 - 12 x + 4 {x}^{2}} \mathrm{dx} = - \frac{1}{4 x + 6} + C$