# Thread: Need Help Starting Problem

1. ## Need Help Starting Problem

Evaluate the integral by making a substitution that converts the integrand to a rational function
$\int\ \frac{e^t}{e^{2t}-4}dx$
can i do the problem this way
$\int\ \frac{e^t}{e^{t^2}-4}dx$
then doin a u sub with $u=e^t$

2. Originally Posted by vinson24
Evaluate the integral by making a substitution that converts the integrand to a rational function
$\int\ \frac{e^t}{e^{2t}-4}dx$
can i do the problem this way
$\int\ \frac{e^t}{e^{t^2}-4}dx$
then doin a u sub with $u=e^t$
$\int {\frac{{du}}
{{u^2 - 4}} = \frac{1}
{4}\int {\left( {\frac{1}
{{u - 2}} - \frac{1}
{{u + 2}}} \right)} du}$

3. Originally Posted by Plato
$\int {\frac{{du}}
{{u^2 - 4}} = \frac{1}
{4}\int {\left( {\frac{1}
{{u - 2}} - \frac{1}
{{u + 2}}} \right)} du}$
How did you get the 1/4?

4. $\frac{1}
{{u - 2}} - \frac{1}
{{u + 2}} = \frac{4}
{{u^2 - 4}}$

5. Originally Posted by vinson24
Evaluate the integral by making a substitution that converts the integrand to a rational function
$\int\ \frac{e^t}{e^{2t}-4}dx$
can i do the problem this way
$\int\ \frac{e^t}{e^{t^2}-4}dx$
then doin a u sub with $u=e^t$
$\int {\frac{e^y}{-4+e^{2t}}} dt$

substitute $u = e^t$ and $du = e^{t} dt$ :

= $\int {\frac{1}{u^{2}-4}} du$

= $\frac{-1}{2} tanh^{-1}\frac{u}{2}+C$

Substitute back for $u = e^t$:

= $\frac{-1}{2}tanh^{-1}\frac{e^t}{2}+C$

Which is equivalent for t values to:

= $\frac{1}{4} log(e^{t}-2)-\frac{1}{4} log(e^{t}+2)+C$