Transform a square-root integral and reverse an exponential chain

From $u=\sqrt{x+1}$ obtain $x=u^2-1$ and $dx=2u\,du$. The integral becomes $2\int(u^2-1)^2du=2\int(u^4-2u^2+1)du=\frac25u^5-\frac43u^3+2u+C$. Substituting back gives $\frac25(x+1)^{5/2}-\frac43(x+1)^{3/2}+2\sqrt{x+1}+C$.

Insert the needed factor 5: $\frac15\int5e^{5x}(e^{5x}-e)^{-4}dx=-\frac1{15}(e^{5x}-e)^{-3}+C$. The original pole is $x=1/5$.