find the indefinite integral by using partial fractions

Question

∫x^2+3x-7⁄(2x+3)(x+1)^2dx

Answer 1

∫(x²-3x7)/(2x+3)(x+1)²dx

After performing the partial fraction decomposition we get

∫(x²-3x-7)/(2x+3)(x+1)²dx = ∫[ 19/(x+1) - 37/(2x+3) - 9/ (x+1)²] d x

= 9/(x+1)+ 19ln | x+1| - (37/2) ln| 2x +3| +λ, where λ is constant