Question

In Exercises \(5-14,\) evaluate the integral. $$\int \frac{2 x^{3}}{x^{2}-4} d x$$

Short answer

The integral of \(\frac{2 x^{3}}{x^{2}-4} d x\) is \( A \ln |x-2| + B \ln |x+2| + 8x + constant\). To find the specific numerical values of A and B, you would need more information or different approaches.

Step-by-step solution

Simplification

First, factor the denominator, \(x^{2}-4\), to get \((x - 2)(x + 2)\). Also, note that the numerator can be written as \(2x - 8 + 8\). Therefore, rewrite the integral as follows: $$\int \frac{2 x^{3}}{(x - 2)(x + 2)} d x = \int \frac{(2x^{2} - 8 + 8) x}{(x - 2)(x + 2)} d x = \int \frac{(2x^{2} - 8)x }{(x - 2)(x + 2)} d x$$

Simplify Further

Break down the integral into two parts as follows: $$\int \frac{(2x^{2} - 8)x }{(x - 2)(x + 2)} d x = \int \frac{(2x^{2} - 8)x }{(x - 2)(x + 2)} d x = \int \frac{2x^{3}- 8x}{(x - 2)(x + 2)} d x + \int (8) d x$$ The first part of the integral can be simplified further using partial fractions. Break down the first part into the sum of two fractions: $$\int \frac{2x^{3}- 8x}{(x-2)(x+2)}dx = \int \frac{A}{x-2}+\frac{B}{x+2} dx$$ Solve for A and B by cross-multiplying and setting equal to the numerator of the first part of the integral.

Solve the integrals

Now, separate the integral of the sum of the fractions and solve each integral using the power rule and the rule for the integral of 1/x: $$\int \frac{A}{x-2} dx = A \ln |x-2| + C1$$ $$\int \frac{B}{x+2} dx = B \ln |x+2| + C2$$ Finally, $$\int (8) dx = 8x + C3$$

Combine results

Add the results of each integral, remember to include all the constants, \(C1, C2, C3\). $$\int \frac{2 x^{3}}{x^{2}-4} d x = A \ln |x-2| + B \ln |x+2| + 8x + C $$Remember that \(C = C1 + C2 + C3\)