Evaluate the improper integral or state that it is divergent.
-$$\int _ { 0 } ^ { \infty } \frac { 2 d x } { 64 + x ^ { 2 } }$$
A) $\pi + 8$
B) $\frac { \pi } { 64 }$
C) 0
D) $\frac { \pi } { 8 }$

Question

Quizplus · Accepted Answer

We can solve this using a substitution. Let $u = \frac{x}{8}$, then $du = \frac{1}{8} dx$ and the integral becomes:
$$\int_0^\infty \frac{2dx}{64 + x^2} = \int_0^\infty \frac{16}{64 + 64u^2}du = \frac{1}{4} \int_0^\infty \frac{1}{1 + u^2}du$$
Now, we have a well-known integral:
$$\int \frac{1}{1 + u^2}du = \arctan u + C$$
So, our integral becomes:
$$\frac{1}{4} \int_0^\infty \frac{1}{1 + u^2}du = \frac{1}{4} \left[\arctan(u)\right]_0^\infty = \frac{1}{4} \left(\frac{\pi}{2} - 0\right) = \frac{\pi}{8}$$
Therefore, the correct answer is D.

Evaluate the Improper Integral or State That It Is Divergent ∫0∞2dx64+x2\int _ { 0 } ^ { \infty } \frac { 2 d x } { 64 + x ^ { 2 } }∫0∞​64+x22dx​

Evaluate the Improper Integral or State That It Is Divergent $\int _ { 0 } ^ { \infty } \frac { 2 d x } { 64 + x ^ { 2 } }$