We recognize that the denominator of the fraction is a
perfect square:
16x^2+24x+9 =
(4x+3)^2
We'll re-write the
integral:
Int f(x)dx = Int
dx/(4x+3)^2
We'll change the
variable.
We'll substitute 4x+3 by
t.
4x+3 = t
We'll
differentiate both sides:
(4x+3)'dx =
dt
So, 4dx = dt
dx =
dt/4
We'll re-write the integral in the variable
t:
Int dx/(4x+3)^2= Int
dt/4t^2
Int dt/4t^2 = Int
[t^(-2)/4]*dt
Int [t^(-2)/4]*dt = t^(-2+1)/4*(-2+1) + C =
t^(-1)/-4 + C = -1/4t + C
But t =
4x+3
Int dx/(4x+3)^2 = -1/4(4x+3) +
C
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