Integration by change the variable

Question

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

Answer 1

Continuing your work, we will obtain \begin{align} \int\sqrt{1+e^x}\ dx&=2\int\ du+2\int\frac1{u^2-1}\ du\\ &=2u+\int\left[\frac1{u-1}-\frac1{u+1}\right]\ du\\ &=2u+\ln|u-1|-\ln|u+1|+C\\ &=2u+\ln\left|\frac{u-1}{u+1}\right|+C\\ &=2\sqrt{1+e^x}+\ln\left|\frac{\sqrt{1+e^x}-1}{\sqrt{1+e^x}+1}\right|+C. \end{align}

Answer 2

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

Answer 3

You've done just fine thus far. For the second integral, if you don't want to use partial fraction decomposition, which is likely the easiest route, then the correct trigonometric substitution would be $$u = \sec \theta\implies du = \sec\theta\tan\theta \,d\theta$$