Chapter 6.5, Problem 7E

To determine

To find: The value of the integral.

Answer to Problem 7E

The integral is:

  $x^2+4\ln \left|x^2-4\right|+C$

Explanation of Solution

Given information:

The integral is: $\int \frac{2x^3}{x^2-4}dx$

Calculation:

The given integral is: $\int \frac{2x^3}{x^2-4}dx$

use polynomial long division to rebuild the function since the degree in the numerator is higher than the degree in the denominator. This means that the quotient is $2x$ and the remainder is $8x$ .

  $\int \frac{2x^3}{x^2-4}dx=\int \left(2x+\frac{8x}{x^2-4}\right)dx$

The integral should be divided into two integrals.

  $\int 2xdx+\int \frac{8x}{x^2-4}dx$

Use the power rule to integrate the first integral. Since the sum of two constants is still a constant and so don't need to write it for this anti derivative, so only need one $+C$ in the final answer.

  $\begin{array}{c}\int \frac{8x}{x^2-4}dx=\int \frac{8}{u}\left(\frac{1}{2}du\right)\\ =\int \frac{4}{u}du\\ =4\ln \left|u\right|+C\\ =4\ln \left|x^2-4\right|+C\end{array}$

For the answer, combine the two anti derivates.

  $\int 2xdx+\int \frac{8x}{x^2-4}dx=x^2+4\ln \left|x^2-4\right|+C$

Therefore the required value of the given integral $\int \frac{2x^3}{x^2-4}dx$ is

  $x^2+4\ln \left|x^2-4\right|+C$