# Thread: Integral of an exponential

1. ## Integral of an exponential

I have been trying to evaluate:

$\displaystyle \int \frac{1}{\sqrt(e^x+1)} dx$

I'd appreciate it if anyone could go through this step by step? My teacher is dropping this on us the day before the test unfortunately. Thanks.

2. Put $\displaystyle t=\sqrt{1+e^{x}}\implies \frac{2}{t^{2}-1}\,dt=\frac{dx}{\sqrt{1+e^{x}}}.$

4. after that t= e^x+1 :the substitution. u should get an integral to solve by-parts.

Another approach . . .

$\displaystyle \int \frac{dx}{\sqrt{e^x+1}}$

Let: .$\displaystyle u \:=\:\sqrt{e^x+1} \quad\Rightarrow\quad u^2 \:=\:e^x+1 \quad\Rightarrow\quad e^x \:=\:u^2-1 \quad\Rightarrow\quad x \:=\:\ln(u^2-1)$

. . Then: .$\displaystyle dx \:=\:\frac{2u\,du}{u^2-1}$

Substitute: .$\displaystyle \int\frac{\frac{2u\,du}{u^2-1}}{u} \;=\;2\int\frac{du}{u^2-1}$

At this point, you can use Partial Fractions or a standard formula

. . and get: .$\displaystyle \ln\left|\frac{u-1}{u+1}\right| + C$

Back-substitute: . $\displaystyle \ln\left|\frac{\sqrt{e^x+1} - 1}{\sqrt{e^x+1} + 1}\right| + C$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

If you want to impress/surprise/terrify your teacher, simplify . . .

Multiply the fraction by: $\displaystyle \frac{\sqrt{e^x+1} - 1}{\sqrt{e^x+1}-1}$

. . $\displaystyle \frac{\sqrt{e^x+1} - 1}{\sqrt{e^x+1} + 1}\cdot{\color{blue}\frac{\sqrt{e^x+1} - 1}{\sqrt{e^x+1} - 1}} \;=\; \frac{(\sqrt{e^x+1} - 1)^2}{(e^x+1) - 1} \;=\; \frac{(\sqrt{e^x+1} - 1)^2}{e^x}$

And the answer becomes: .$\displaystyle \ln\left(\frac{(\sqrt{e^x+1} - 1)^2}{e^x}\right) + C$

. . . $\displaystyle =\;\ln\left((\sqrt{e^x+1} - 1)^2\right) - \ln(e^x) + C \;=\;2\ln\left(\sqrt{e^x+1}-1\right) - x + C$

6. actually, you used my substitution as well.