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

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Answer 1 · 2017-05-25T11:37:24

#-1/(4x+6)+C#

Let's write the question formatted:

#int 1/(9-12x+4x^2) dx#

First off, we can simplify the denominator

#1/(9-12x+4x^2)=1/(2x+3)^2#

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

#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)) ?