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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