Evaluate ∫dxx+1\int \frac{dx}{x+1}∫x+1dx for x>−1x > -1x>−1.
ln(x+1)+C\ln(x+1) + Cln(x+1)+C
1(x+1)2+C\frac{1}{(x+1)^2} + C(x+1)21+C
ln∣x+1∣+C\ln|x+1| + Cln∣x+1∣+C
exp(x+1)+C\text{exp}(x+1) + Cexp(x+1)+C