Trying to evaluate : $\int \frac{e^{x^2}+2xe^{x}}{e^{x^2}+e^x} dx$

Question

Here's my approach :

$$\int \frac{e^{x^2}+2xe^{x}}{e^{x^2}+e^x} dx = \int\sum\limits_{n=0}^{\infty} \frac{\frac{(x^2)^n}{n!}+\frac{2x^{n+1}}{n!}}{\frac{(x^2)^n}{n!}+\frac{x^n}{n!}} dx = \int\sum\limits_{n=0}^{\infty} \frac{x^{2n}+2x^{n+1}}{x^{2n}+x^{n}} dx$$ Now interchanging the summation idk if it is justified.

$$ = \sum\limits_{n=0}^{\infty}\int \frac{x^{2n}+2x^{n+1}}{x^{2n}+x^{n+1}} dx = \sum\limits_{n=0}^{\infty}\int \frac{x^n+2x}{x^n+1} = \sum\limits_{n=0}^{\infty}\int 1 + \frac{2x-1}{x^n+1} dx = \sum\limits_{n=0}^{\infty}\left[\int 1 dx + \int \frac{2x}{x^n+1} dx - \frac{1}{x^n+1} dx\right] $$

The first one is pretty evident, second and third i'm not so sure.

Does this integral even have an elementary anti derivative? If so, please provide a full solution.

Answer 1

Here is an alternative way

$\displaystyle\int \frac{e^{x^2}+2xe^{x}}{e^{x^2}+e^x} dx=\int \frac{e^{x^2}+2xe^{x}+e^x-e^x}{e^{x^2}+e^x} dx=$

$\displaystyle x+\int \frac{(2x-1)e^{x}}{e^{x^2}+e^x} dx=x+\int \frac{(2x-1)}{e^{x^2-x}+1} dx=x+\int\frac{du}{e^u-1}$

Can you now continue?

Answer 2

Using Mathematica, we can obtain via:

Integrate[(Exp[x^2] + 2 x*Exp[x])/(Exp[x^2] + Exp[x]), x]

an exact form:

$$\int \frac{e^{x^2}+2 x e^x}{e^{x^2}+e^x} \, dx=x^2-\log \left(e^{x^2}+e^x\right)+x$$

the plot looks as follows:

Behind the scenes it shows the transformation:

$$\int 1+\frac{e^x(2x-1)}{e^x+e^{x^2}} \, dx=x+\int \frac{e^x(2x-1)}{e^x+e^{x^2}} \, dx=x+\int \frac{2e^xx}{e^x+e^{x^2}}-\frac{e^x}{e^x+e^{x^2}} \, dx=x+\int \frac{2e^xx}{e^x+e^{x^2}}\, dx-\int \frac{e^x}{e^x+e^{x^2}} \, dx$$