Decide whether the integral converges. If the integral converges, compute its value. $\int _ { - 9 } ^ { 9 } \left( \frac { 1 } { ( x + 8 ) ^ { \frac { 1 } { 5 } } } \right) d x$
A) $\left( 17 ^ { \frac { 1 } { 5 } } - 1 \right)$
B) $\frac { 5 } { 4 } \left( 17 ^ { \frac { 4 } { 5 } } - 1 \right)$
C) $\left( 17 ^ { \frac { 2 } { 5 } } - 1 \right)$
D) $- \frac { 5 } { 4 } \left( 17 ^ { - \frac { 4 } { 5 } } + 1 \right)$
E) diverges

Question

Decide whether the integral converges. If the integral converges, compute its value.   $\int _ { - 9 } ^ { 9 } \left( \frac { 1 } { ( x + 8 ) ^ { \frac { 1 } { 5 } } } \right) d x$  
A) $\left( 17 ^ { \frac { 1 } { 5 } } - 1 \right)$
B) $\frac { 5 } { 4 } \left( 17 ^ { \frac { 4 } { 5 } } - 1 \right)$
C) $\left( 17 ^ { \frac { 2 } { 5 } } - 1 \right)$
D) $- \frac { 5 } { 4 } \left( 17 ^ { - \frac { 4 } { 5 } } + 1 \right)$
E) diverges

Quizplus · Accepted Answer

The integral converges because the function $ \frac { 1 } { ( x + 8 ) ^ { \frac { 1 } { 5 } } } $ is continuous on the interval $[-9, 9]$ except at $x = -8$, but the singularity at $x = -8$ is not strong enough to cause divergence. To compute the integral, use the substitution $u = x + 8$, which gives $du = dx$, and the integral becomes $\int _ { - 1 } ^ { 17 } u ^ { - \frac { 1 } { 5 } } du$. This can be integrated directly to $\frac { 5 } { 4 } u ^ { \frac { 4 } { 5 } }$ evaluated from $-1$ to $17$, resulting in $\frac { 5 } { 4 } \left( 17 ^ { \frac { 4 } { 5 } } - 1 \right)$.

Decide Whether the Integral Converges ∫−99(1(x+8)15)dx\int _ { - 9 } ^ { 9 } \left( \frac { 1 } { ( x + 8 ) ^ { \frac { 1 } { 5 } } } \right) d x∫−99​((x+8)51​1​)dx

Decide Whether the Integral Converges $\int _ { - 9 } ^ { 9 } \left( \frac { 1 } { ( x + 8 ) ^ { \frac { 1 } { 5 } } } \right) d x$