Question

evaluate the following integral or state that it diverges.
int_{ln 1/4}^{infty}\frac{e^{x}}{16e^{2x}+1}dx
select the correct choice and, if necessary, fill in the answer box to complete your choice.
a. the improper integral converges and int_{ln 1/4}^{infty}\frac{e^{x}}{16e^{2x}+1}dx=square .
b. the improper integral diverges.

Explanation:

Step1: Substitute $u = e^{x}$

$du=e^{x}dx$. When $x = \ln\frac{1}{4}$, $u=\frac{1}{4}$; as $x
ightarrow\infty$, $u
ightarrow\infty$. The integral $\int_{\ln\frac{1}{4}}^{\infty}\frac{e^{x}}{16e^{2x}+1}dx$ becomes $\int_{\frac{1}{4}}^{\infty}\frac{du}{16u^{2}+1}$.

Step2: Rewrite the integrand

We know that $\int\frac{du}{16u^{2}+1}=\int\frac{du}{(4u)^{2}+1}$. Let $v = 4u$, then $dv=4du$ and $du=\frac{1}{4}dv$. The integral becomes $\frac{1}{4}\int_{\ 1}^{\infty}\frac{dv}{v^{2}+1}$.

Step3: Recall the antiderivative

The antiderivative of $\frac{1}{v^{2}+1}$ is $\arctan(v)$. So, $\frac{1}{4}\int_{1}^{\infty}\frac{dv}{v^{2}+1}=\frac{1}{4}\lim_{b
ightarrow\infty}(\arctan(v)\big|_{1}^{b})$.

Step4: Evaluate the limit

$\frac{1}{4}\lim_{b
ightarrow\infty}(\arctan(b)-\arctan(1))=\frac{1}{4}(\frac{\pi}{2}-\frac{\pi}{4})$.

Step5: Simplify the result

$\frac{1}{4}\times\frac{\pi}{4}=\frac{\pi}{16}$.

Answer:

A. The improper integral converges and $\int_{\ln\frac{1}{4}}^{\infty}\frac{e^{x}}{16e^{2x}+1}dx=\frac{\pi}{16}$