Chapter 5: Q. 28 (page 441)

Calculate each of the integrals in Exercises 17–46. For some integrals you may need to use polynomial long division, partial fractions, factoring or expanding, or the method of completing the square.
$\int \frac{3 x^{3}}{x^{3} - 1} d x$

Short Answer

Expert verified

The value of integral is $3 x + \ln |x - 1| - \frac{1}{2} \ln (x^{2} + x + 1) - \sqrt{3} \tan^{- 1} (\frac{2 x + 1}{\sqrt{3}}) + C$ .

Step by step solution

Step 1. Given Information.

The given integral is $\int \frac{3 x^{3}}{x^{3} - 1} d x$ .

Step 2. Calculation.

$\int \frac{3 x^{3}}{x^{3} - 1} d x$

Rewrite the fraction

localid="1652771910496" $\int \frac{3 x^{3}}{x^{3} - 1} d x = \int (3 + \frac{3}{x^{3} - 1}) d x = \int 3 d x + \int \frac{3}{x^{3} - 1} d x = 3 x + 3 \int \frac{1}{(x - 1) (x^{2} + x + 1)} d x = 3 x + 3 \int \frac{1}{3 (x - 1)} d x + 3 \int \frac{(- x - 2)}{3 (x^{2} + x + 1)} d x = 3 x + \ln |x - 1| - \frac{3}{3} \int (\frac{x + 2}{x^{2} + x + 1}) d x = 3 x + \ln |x - 1| - [\frac{(2 x + 1)}{2 (x^{2} + x + 1)} + \frac{3}{2 (x^{2} + x + 1)}] d x = 3 x + \ln |x - 1| - \frac{1}{2} \ln (|x^{2} + x + 1|) - \frac{3}{2} \int \frac{1}{x^{2} + x + 1} d x = 3 x + \ln |x - 1| - \frac{1}{2} \ln (|x^{2} + x + 1|) - \frac{3}{2} \int \frac{1}{{(x + \frac{1}{x})}^{2} + \frac{3}{4}} d x$

Step 3. Calculation.

Let

$u = \frac{2 x + 1}{\sqrt{3}} d u = \frac{2}{\sqrt{3}} d x d x = \frac{\sqrt{3}}{2} d u$

localid="1652771882551" $3 x + \ln |x - 1| - \frac{1}{2} \ln (|x^{2} + x + 1|) - \frac{3}{2} \int \frac{1}{{(x + \frac{1}{x})}^{2} + \frac{3}{4}} d x = 3 x + \ln |x - 1| - \frac{1}{2} \ln (|x^{2} + x + 1|) - \frac{3}{2} \int \frac{\sqrt{3}}{2 (\frac{3 u^{2}}{4} + \frac{3}{4})} d u = 3 x + \ln |x - 1| - \frac{1}{2} \ln (|x^{2} + x + 1|) - \frac{3 \sqrt{3}}{4} \int (\frac{d u}{3 u^{2} + 3}) = 3 x + \ln |x - 1| - \frac{1}{2} \ln (|x^{2} + x + 1|) - \sqrt{3} \tan^{- 1} (u) = 3 x + \ln |x - 1| - \frac{1}{2} \ln (|x^{2} + x + 1|) - \sqrt{3} \tan^{- 1} (\frac{2 x + 1}{\sqrt{3}}) + C$

Step 4. Conclusion.

The value of integral is $3 x + \ln |x - 1| - \frac{1}{2} \ln (|x^{2} + x + 1|) - \sqrt{3} \tan^{- 1} (\frac{2 x + 1}{\sqrt{3}}) + C$ .

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Calculate each of the integrals in Exercises 17–46. For some integrals you may need to use polynomial long division, partial fractions, factoring or expanding, or the method of completing the square.
$\int \frac{3 x^{3}}{x^{3} - 1} d x$

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Calculation.

Step 3. Calculation.

Step 4. Conclusion.

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Calculate each of the integrals in Exercises 17–46. For some integrals you may need to use polynomial long division, partial fractions, factoring or expanding, or the method of completing the square.∫3x3x3-1dx

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Calculation.

Step 3. Calculation.

Step 4. Conclusion.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Pure Maths

Calculus

Mechanics Maths

Probability and Statistics

Statistics

Geometry

Study anywhere. Anytime. Across all devices.

Calculate each of the integrals in Exercises 17–46. For some integrals you may need to use polynomial long division, partial fractions, factoring or expanding, or the method of completing the square.
$\int \frac{3 x^{3}}{x^{3} - 1} d x$