Integration by parts or using partial fractions is overkill. Both can be done with simple substitutions.
For the first integral, take y = x + 1 and dy = dx.
[tex]\displaystyle \int \frac2{(x+1)^2} \, dx = \int \frac2{y^2} \, dy = -\frac2y + C \\\\ = -\frac2{x+1} + C[/tex]
For the second integral, use the same substitution.
[tex]\displaystyle \int \frac{x-1}{(x+1)^2} \, dx = \int \frac{(y-1)-1}{y^2} \, dy = \int \left(\frac1y - \frac2{y^2}\right) \, dy = \ln|y| + \frac2y + C \\\\ = \ln|x+1| + \frac2{x+1} + C[/tex]
