Evaluate the indefinite integral of the fraction 1/(16x^2+24x+9)?
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We have to find the integral of 1/(16x^2+24x+9)
1/(16x^2+24x+9)
=> 1/(4x+3)^2
let 4x + 3 = u
=> du/dx = 4
=> dx = du/4
Int[ 1/(16x^2+24x+9) dx]
=> Int [ 1/u^2 du/4]
=> (1/4) Int [ u^-2 du]
=> (1/4) u^-1/-1
=> (-1/4u) + C
replace u = 4x + 3
=> -1/(16x + 12) + C
The required integral is -1/(16x + 12) + C
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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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