The integral $\int \frac { 1 } { \sqrt { 8 - 2 x - x ^ { 2 } } } d x$ is

A) $\sin ^ { - 1 } \left( \frac { x + 1 } { 3 } \right) + C$
B) $- \sin ^ { - 1 } \left( \frac { x + 1 } { 3 } \right) + C$
C) $\frac { \sin ^ { - 1 } \left( \frac { x + 1 } { 3 } \right) } { 3 } + C$
D) $- \frac { \sin ^ { - 1 } \left( \frac { x + 1 } { 3 } \right) } { 3 } + C$
E) $\frac { \sin ^ { - 1 } \left( \frac { x - 1 } { 3 } \right) } { 9 } + C$

Question

The integral $\int \frac { 1 } { \sqrt { 8 - 2 x - x ^ { 2 } } } d x$ is

A) $\sin ^ { - 1 } \left( \frac { x + 1 } { 3 } \right) + C$
B) $- \sin ^ { - 1 } \left( \frac { x + 1 } { 3 } \right) + C$
C) $\frac { \sin ^ { - 1 } \left( \frac { x + 1 } { 3 } \right) } { 3 } + C$
D) $- \frac { \sin ^ { - 1 } \left( \frac { x + 1 } { 3 } \right) } { 3 } + C$
E) $\frac { \sin ^ { - 1 } \left( \frac { x - 1 } { 3 } \right) } { 9 } + C$

Quizplus · Accepted Answer

To solve the integral $\int \frac { 1 } { \sqrt { 8 - 2 x - x ^ { 2 } } } d x$, we first complete the square in the denominator to get it in the form of $\sqrt{a^2 - (x+b)^2}$. The expression $8 - 2x - x^2$ can be rewritten as $- (x^2 + 2x - 8)$, which becomes $- [(x + 1)^2 - 9]$ after completing the square. This gives us $\sqrt{9 - (x + 1)^2}$, which matches the form needed to use the integral formula for $\sin^{-1}$. The correct answer is $\sin^{-1}\left(\frac{x + 1}{3}\right) + C$, since $a = 3$ in this context, and the variable part inside the $\sin^{-1}$ function is $\frac{x + 1}{3}$.

The Integral ∫18−2x−x2dx\int \frac { 1 } { \sqrt { 8 - 2 x - x ^ { 2 } } } d x∫8−2x−x2​1​dx

The Integral $\int \frac { 1 } { \sqrt { 8 - 2 x - x ^ { 2 } } } d x$