Evaluate the integral by first performing long division on the integrand and then writing the proper fraction as a sum of partial fractions.
-$$\int \frac { 5 x ^ { 3 } + 7 x ^ { 2 } - 2 x - 4 } { x ^ { 3 } - x ^ { 2 } } d x$$
A) $\frac { 5 } { x } + 6 \ln | x | - \frac { 4 } { x ^ { 2 } } + 6 \ln | x - 1 | + C$
B) $5 x + 6 \ln \left| x ^ { 2 } \right| - \frac { 4 } { x ^ { 2 } } + 6 \ln | x - 1 | + C$
C) $5 x + 7 \ln | x | + \frac { 4 } { x } + 6 \ln | x - 1 | + C$
D) $5 x + 6 \ln | x | - \frac { 4 } { x } + 6 \ln | x - 1 | + C$

Question

Quizplus · Accepted Answer

First, we perform long division and obtain:
$$\frac{5x^3+7x^2-2x-4}{x^3-x^2} = 5+\frac{6x^2-4}{x^3-x^2}.$$
We can rewrite the proper fraction as a sum of partial fractions as follows:
$$\frac{6x^2-4}{x^3-x^2} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x^2}.$$
Multiplying both sides by the denominator and solving for the unknowns $A, B,$ and $C$ yields:
$$A=5, \quad B=6, \quad	ext{and}\quad C=-4.$$
Therefore,
\begin{align*}
\int \frac { 5 x ^ { 3 } + 7 x ^ { 2 } - 2 x - 4 } { x ^ { 3 } - x ^ { 2 } } d x &= \int \left(5+\frac{6x^2-4}{x^3-x^2}ight)dx \
&= 5x + \int\frac{6x^2-4}{x^2(x-1)}dx \
&= 5x + \int\left(\frac{A}{x} + \frac{B}{x-1} + \frac{C}{x^2}ight)dx \
&= 5x + 6\ln|x-1| -\frac{4}{x} + C,
\end{align*}
where we have used the fact that $\int\frac{dx}{x} = \ln|x| + C.$

Evaluate the Integral by First Performing Long Division on the Integrand