Answer
$(x+1) \ln (x+1)-x+c$
Work Step by Step
Integration by parts formula suggests that $\int p'(x) q(x)=p(x) q(x)-\int p(x) q'(x)dx$
$\int \ln (x+1) dx=(x+1) \ln (x+1) -\int (x+1) (\dfrac{1}{(x+1)})dx$
Thus, $(x+1) \ln (x+1)-\int 1 dx=(x+1) \ln (x+1)-x+c$
