$\text { Use partial fractions to find } \int \frac { 1 } { x ^ { 2 } - 9 } d x \text {. }$
A) $\int \frac { 1 } { x ^ { 2 } - 9 } d x = \frac { 1 } { 6 } \ln \left| \frac { x - 3 } { x + 3 } \right| + C$
B) $\int \frac { 1 } { x ^ { 2 } - 9 } d x = \frac { 1 } { 6 ( x - 3 ) } - \frac { 1 } { 6 ( x + 3 ) } + C$
C) $\int \frac { 1 } { x ^ { 2 } - 9 } d x = \frac { 1 } { 6 } \ln \left| \frac { x - 6 } { x + 6 } \right| + C$
D) $\int \frac { 1 } { x ^ { 2 } - 9 } d x = \frac { 1 } { ( x - 3 ) } - \frac { 1 } { ( x + 3 ) } + C$
E)$\int \frac { 1 } { x ^ { 2 } - 9 } d x = \ln \left| \frac { x - 3 } { x + 3 } \right| + C$

Question

$\text { Use partial fractions to find } \int \frac { 1 } { x ^ { 2 } - 9 } d x \text {. }$ 
A) $\int \frac { 1 } { x ^ { 2 } - 9 } d x = \frac { 1 } { 6 } \ln \left| \frac { x - 3 } { x + 3 } \right| + C$
B) $\int \frac { 1 } { x ^ { 2 } - 9 } d x = \frac { 1 } { 6 ( x - 3 ) } - \frac { 1 } { 6 ( x + 3 ) } + C$
C) $\int \frac { 1 } { x ^ { 2 } - 9 } d x = \frac { 1 } { 6 } \ln \left| \frac { x - 6 } { x + 6 } \right| + C$
D) $\int \frac { 1 } { x ^ { 2 } - 9 } d x = \frac { 1 } { ( x - 3 ) } - \frac { 1 } { ( x + 3 ) } + C$
E)$\int \frac { 1 } { x ^ { 2 } - 9 } d x = \ln \left| \frac { x - 3 } { x + 3 } \right| + C$

Quizplus · Accepted Answer

The integral $\int \frac { 1 } { x ^ { 2 } - 9 } d x$ can be solved using partial fractions by expressing $\frac { 1 } { x ^ { 2 } - 9 }$ as $\frac { 1 } { (x-3)(x+3) }$. When decomposed, it leads to $\frac { A } { x-3 } + \frac { B } { x+3 }$, where A and B are constants that can be solved to give $\frac { 1 } { 6(x-3) } - \frac { 1 } { 6(x+3) }$. Integrating this expression leads to $\frac { 1 } { 6 } \ln \left| \frac { x - 3 } { x + 3 } \right| + C$, which matches choice A.

Use partial fractions to find ∫1x2−9dx. \text { Use partial fractions to find } \int \frac { 1 } { x ^ { 2 } - 9 } d x \text {. } Use partial fractions to find ∫x2−91​dx.

$\text { Use partial fractions to find } \int \frac { 1 } { x ^ { 2 } - 9 } d x \text {. }$