Question

evaluate the following integral.
int\frac{e^{4x}}{e^{4x}-5e^{-4x}}dx

Explanation:

Step1: Use substitution

Let $u = e^{4x}-5e^{- 4x}$, then $du=(4e^{4x}+20e^{-4x})dx = 4(e^{4x}+5e^{-4x})dx$.

Step2: Rewrite the integral

We have $\int\frac{e^{4x}}{e^{4x}-5e^{-4x}}dx=\frac{1}{4}\int\frac{4e^{4x}}{e^{4x}-5e^{-4x}}dx$.

Step3: Substitute $u$ and $du$

Substituting $u$ and $du$ into the integral, we get $\frac{1}{4}\int\frac{du}{u}$.

Step4: Integrate with respect to $u$

The integral of $\frac{1}{u}$ with respect to $u$ is $\ln|u|+C$. So $\frac{1}{4}\int\frac{du}{u}=\frac{1}{4}\ln|u|+C$.

Step5: Substitute back $u$

Substituting $u = e^{4x}-5e^{-4x}$ back, we have $\frac{1}{4}\ln|e^{4x}-5e^{-4x}|+C$.

Answer:

$\frac{1}{4}\ln|e^{4x}-5e^{-4x}|+C$