How do you integrate #\int \frac { d x } { ( 2x - 3) ^ { 2} }#?

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Answer 1 · 2017-06-10T14:56:42

The answer is #=-1/(2(2x-3))+C#

We need

#intx^2dx=x^(n+1)/(n+1)+C(n!=-1)#

We perform this integral by substitution.

Let #u=2x-3#, #=>#, ##du=2dx

So,

#int(dx)/(2x-3)^2=1/2int(du)/u^2=1/2intu^-2du#

#=-1/2u^-1#

#=-1/(2(2x-3))+C#