# Evaluate the indefinite integral of the fraction 1/(16x^2+24x+9)?

### 2 Answers | Add Yours

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**

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**