Evaluate the integral by first performing long division on the integrand and then writing the proper fraction as a sum of partial fractions.
-$$\int \frac { x ^ { 4 } } { x ^ { 2 } - 9 } d x$$
A) $\frac { x ^ { 3 } } { 3 } + 9 x + \frac { 27 } { 2 } \ln | x - 9 | - \frac { 27 } { 2 } \ln | x + 9 | + C$
B) $\frac { x ^ { 3 } } { 3 } + 9 x + \frac { 27 } { 2 } \ln | x - 3 | - \frac { 27 } { 2 } \ln | x + 3 | + C$
C) $\frac { x ^ { 3 } } { 3 } + \frac { 9 } { 2 } \ln | x - 3 | - \frac { 9 } { 2 } \ln | x + 3 | + C$
D) $\frac { x ^ { 3 } } { 3 } + 9 x - \frac { 27 } { 2 } \ln | x - 3 | + \frac { 27 } { 2 } \ln | x + 3 | + C$

Question

Quizplus · Accepted Answer

Performing long division on the integrand, we have:
$$\frac{x^4}{x^2-9} = x^2 + 9 + \frac{81}{x^2-9}$$

Now we need to write $\frac{81}{x^2-9}$ as a sum of partial fractions. Let $u=x^2-9$. Then:

$$\frac{81}{x^2-9}=\frac{81}{u}=\frac{A}{u+3}+\frac{B}{u-3}$$

Multiplying both sides by $(u+3)(u-3)$, we have:

$$81=A(u-3)+B(u+3)$$

Plugging in $u=3$, we get $A=9$. Plugging in $u=-3$, we get $B=-9$. Therefore:

$$\frac{81}{x^2-9}=\frac{9}{x+3}-\frac{9}{x-3}$$

Substituting everything back into the original integral, we have:

$$\int \frac { x ^ { 4 } } { x ^ { 2 } - 9 } d x = \int (x^2 + 9 + \frac{9}{x-3}-\frac{9}{x+3}) dx = \frac { x ^ { 3 } } { 3 } + 9 x + 9\ln | x - 3 | - 9\ln | x + 3 | + C$$

Simplifying, we get:

$$\frac { x ^ { 3 } } { 3 } + 9 x + \frac { 27 } { 2 } \ln | x - 3 | - \frac { 27 } { 2 } \ln | x + 3 | + C$$

Therefore, the answer is choice B.

Evaluate the Integral by First Performing Long Division on the Integrand