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

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Answer 1 · 2017-05-25T11:37:24

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

Let's write the question formatted:

$\int \frac{1}{9 - 12 x + 4 x^{2}} d x$ $int 1/(9-12x+4x^2) dx$

First off, we can simplify the denominator

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

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

$\int f (u) \frac{d u}{d x} d x = \int f (u) d u$ $int f(u) (du)/dx dx=int f(u) du$

Let $u=2x+3, \quad (du)/dx=d/dx[2x+3]=2$
We get:

$1/2int 1/u^2 du=1/2intu^(-2)du$

Solve.

$1/2intu^(-2)du=-1/(2u)+C$

Sub $u$ back.

$-1/(2u)+C=-1/(2(2x+3))+C=-1/(4x+6)+C$

$:.$ $int 1/(9-12x+4x^2) dx=-1/(4x+6)+C$

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