# integration by substitution

• Jul 20th 2011, 06:13 AM
Punch
integration by substitution
$\int\frac{1}{e^x+e^{-x}}dx {$ let $u=e^x}$

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

$=\int{1}{u^2+1}du$

$=\frac{1}{2}\int\frac{2}{u^2+1}du$

$=\frac{1}{2}ln|u^2+1|+c$

$=\frac{1}{2}ln|e^{2x}+1|+c$
• Jul 20th 2011, 06:37 AM
Prove It
Re: integration by substitution
It's a good idea to get a common denominator...

\displaystyle \begin{align*}\int{\frac{1}{e^x + e^{-x}}\,dx} &= \int{\frac{1}{e^x + \frac{1}{e^x}}\,dx} \\ &= \int{\frac{1}{\frac{e^{2x} + 1}{e^x}}\,dx} \\ &= \int{\frac{e^x}{e^{2x} + 1}\,dx} \end{align*}

Now make the substitution $\displaystyle u = e^x \implies du = e^x\,dx$ and the integral becomes

\displaystyle \begin{align*} \int{\frac{1}{u^2 + 1}\,du} &= \arctan{u} + C \\ &= \arctan{\left(e^x\right)} + C\end{align*}
• Jul 20th 2011, 07:43 AM
Quacky
Re: integration by substitution
Quote:

Originally Posted by Punch
$\int\frac{1}{e^x+e^{-x}}dx {$ let $u=e^x}$

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

$=\int{1}{u^2+1}du$

$=\frac{1}{2}\int\frac{2}{u^2+1}du$

Here is your error. The integration here is not correct. You would need something in terms of 'u' on the numerator to be able to do something along these lines.

$=\frac{1}{2}ln|u^2+1|+c$

$=\frac{1}{2}ln|e^{2x}+1|+c$

Editted.