Integration by change the variable 
Let, $\int_{-1}^1\sqrt{1+e^x}\operatorname{dx}$. Write as an integral of a rational function and compute it.
Suggest: change the variable in order to eliminate the square root.

My work was:
Let $u^2=1+e^x$, so $u=\sqrt{1+e^x}$. One also have $e^x=u^2-1$.
Then one got $\operatorname{du}=\frac{e^x}{2\sqrt{1+e^x}}\operatorname{dx}$ and so $\operatorname{dx}=\frac{2\sqrt{1+e^x}}{e^x}\operatorname{du}$.
Now substituting:
$$\int_{-1}^1\sqrt{1+e^x}\operatorname{dx}=\int_{-1}^1 \sqrt{u^2}\frac{2\sqrt{1+e^x}}{e^x}\operatorname{du}=\int_{-1}^1u\frac{2\sqrt{u^2}}{u^2-1}\operatorname{du}=$$
$$=2\int_{-1}^1\frac{u^2}{u^2-1}\operatorname{du}=2\int_{-1}^11+\frac{1}{u^2-1}\operatorname{du}=$$
$$=2\int_{-1}^11 \operatorname{du}+2\int_{-1}^1\frac{1}{u^2-1}\operatorname{du}$$
Is this thought right?
And what is the second integral? Is not the $\arctan$! Thanks

An easy way to do the second integral is to use the fact that $\frac{1}{u^2 - 1} = \frac{1}{2}(\frac{1}{u-1} - \frac{1}{u+1})$
