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...

## See

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

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**