Is my answer correct for indefinite integral $\int \frac{xe^x}{\sqrt{1+e^x}} dx$

Question

Question: $$\int \frac{xe^x}{\sqrt{1+e^x}} dx$$

If my answer given below is correct, is there an easier way of computing this integral? I feel like this took too long to solve.

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Answer 1

Let $t=e^x$ to rewrite the integral,

$$I=\int \frac{xe^x}{\sqrt{1+e^x}}dx=\int \frac{\ln t}{\sqrt{1+t}}dt$$ Integrate by parts,

$$I = 2\sqrt{1+t}\ln t - 2\int \frac{\sqrt{1+t}}{t}dt\tag 1$$

Then, with $u=\sqrt{1+t}>1$,

$$\int \frac{\sqrt{1+t}}{t}dt=\int \frac{2u^2}{u^2-1}du=2u +\int \frac2{u^2-1}du =2u+\ln\frac{u-1}{u+1}$$

Plug into (1) to obtain

$$I = 2\sqrt{1+t}\ln t - 4u + 2\ln\frac{u-1}{u+1} + C$$ $$ = 2x\sqrt{1+e^x}- 4\sqrt{1+e^x} + 2\ln\frac{\sqrt{1+e^x} +1}{\sqrt{1+e^x} -1} + C$$