Here the denominator is already factored. 

So let,

∫ (2x+1)/(x+1)(x-2)dx

= \(\frac{A}{x+1}+\frac{B}{x-2}\) ...(i)

⇒ \(\int\frac{2x+1}{(x+1)(x-2)}\) = \(\frac{A(x-2)+B(x+1)}{(x+1)(x-2)}\)

⇒ 2x + 1 = A(x – 2) + B(x + 1)……(ii)

We need to solve for A and B. 

One way to do this is to pick values for x which will cancel each variable. 

Put x = 2 in the above equation, we get

⇒ 2(2) + 1 = A(2 – 2) + B(2 + 1)

⇒ 3B = 5

⇒ B = \(\frac{5}{3}\)

Now 

Put x = – 1 in equation (ii), we get 

⇒ 2( – 1) + 1 = A(( – 1) – 2) + B(( – 1) + 1) 

⇒ – 3A = – 1

⇒ A = \(\frac{1}{3}\)

We put the values of A and B values back into our partial fractions in equation (i) and replace this as the integrand. 

We get,

\(\int[\frac{A}{x+1}+\frac{B}{x-2}]\) dx

⇒ \(\int[\frac{\frac{1}{3}}{x+1}+\frac{\frac{5}{3}}{x-2}]\) dx

Split up the integral,

⇒ \(\frac{1}{3}\)\(\int[\frac{1}{x+1}]\)dx + \(\frac{5}{3}\)\(\int[\frac{1}{x-2}]\)dx

Let substitute u = x + 1 

⇒ du = dx and z = x – 2 

⇒ dz = dx, 

So the above equation becomes,

⇒ \(\frac{1}{3}\)\(\int[\frac{1}{u}]\)du + \(\frac{5}{3}\)\(\int[\frac{1}{z}]\)dz

On integrating we get,

⇒ \(\frac{1}{3}\)log|u| + \(\frac{5}{3}\)log|z| + c

Substituting back, we get

⇒ \(\frac{1}{3}\)log|x + 1| + \(\frac{5}{3}\)log|x - 2| + c

Note : the absolute value signs account for the domain of the natural log function (x>0). 

Hence,

\(\int\frac{2x+1}{(x+1)(x-2)}\) dx = \(\frac{1}{3}\)log|x + 1| + \(\frac{5}{3}\)log|x - 2| + c