Question

Evaluate the integral (2x +3)/(x+1)^2 dx from 0 to 1 with integration by parts.

Answer 1

\[\frac{2x+3}{(x+1)^2}=\frac{x+1}{(x+1)^2}+\frac{x+3}{(x+1)^2}\] might be a good first step

Answer 2

Should it not be:
\[\frac{ 2x }{ (x+1)^{2} } + \frac{ 3 }{ (x+1)^{2} }\]

Answer 3

that would work too

Answer 4

alternative
u can use int by u-sub

let u = x+1 --------> x = u - 1
du = dx

see ur integral becomes
int (2(u-1)+3)/u^2 du
= int (2u + 1)/u^2 du
= int (2/u + u^-2) du

(i miss the limits for while)

Answer 5

that should be easier for u, @Prof :)

Answer 6

btw, i didnt read above u want solving by parts
sorry :)