Integrate the following $\int\frac{x+1}{x(1+xe^x)^2}dx$

Question

I tried to multiply the numerator and denominator by $e^{-2x}$

Getting the following simplification $\int\frac{e^{-2x}(x+1)}{x(e^{-x}+x)^2}dx$

My next step is as follow $\int\frac{e^{-2x}(x+1)}{x^3(\frac{e^{-x}}{x}+1)^2})dx$ but could not proceed from here

score 5 · Accepted Answer · answered Apr 20 '19 at 11:02

5

Hint: let $$u=1+xe^x\rightarrow xe^x=u-1$$ $$dx=\frac{du}{e^x(1+x)}$$ substitute them to get $$\int\frac{1}{u^2(u-1)}du$$

answered Apr 20 '19 at 11:02

E.H.E

score 2 · Answer 2 · answered Apr 20 '19 at 10:44

2

Hint: Write your integrand in the form$$-\frac{e^x x+e^x}{e^x x+1}-\frac{e^x x+e^x}{\left(e^x x+1\right)^2}+\frac{1}{x}+1$$

answered Apr 20 '19 at 10:44

2 Answers2