Evaluating the Integral (int frac{3x^7 dx}{x^3 - 1}) and Using Partial Fractions

Understanding how to evaluate integrals, particularly those involving rational functions, is a fundamental skill in calculus. In this article, we will explore the evaluation of the integral (int frac{3x^7 dx}{x^3 - 1}) using the method of partial fractions. This technique is not only useful for solving integrals but also for understanding the structure of rational functions.

Step-by-Step Solution

The integrand (frac{3x^7}{x^3 - 1}) can be rewritten using partial fractions. The denominator (x^3 - 1) can be factored as ((x - 1)(x^2 x 1)). We express the integrand as:

[frac{3x^7}{x^3 - 1} frac{A}{x - 1} frac{Bx C}{x^2 x 1}]

Multiplying both sides by the denominator (x^3 - 1), we have:

[3x^7 A(x^2 x 1) (Bx C)(x - 1)]

Expanding and combining like terms, we get:

[3x^7 Ax^2 Ax A Bx^2 - Bx Cx - C]

[3x^7 (A B)x^2 (A - B C)x (A - C)]

By comparing the coefficients of (x^7), (x^2), (x), and the constant terms on both sides, we can solve for (A), (B), and (C).

Comparing Coefficients

For (x^7) terms: (0 0) For (x^2) terms: (A B 0) For (x) terms: (A - B C 0) For constant terms: (A - C 7)

Solving these equations, we obtain:

From (A B 0), we get (B -A). From (A - B C 0), substituting (B -A), we get (A A C 0) or (C -2A). From (A - C 7), substituting (C -2A), we get (A 2A 7) or (3A 7), thus (A frac{7}{3}). Thus, (B -frac{7}{3}) and (C -frac{14}{3}).

Substituting (A), (B), and (C) into the integral, we have:

[int frac{3x^7 dx}{x^3 - 1} frac{1}{3} left[ int frac{7 dx}{x - 1} - int frac{7x 14}{x^2 x 1} dx right]]

The first integral is straightforward:

[int frac{7 dx}{x - 1} 7 ln |x - 1| C_1]

For the second integral, we use the standard forms for the logarithm and the inverse tangent function:

[int frac{7x 14}{x^2 x 1} dx 7 int frac{2x 1 3}{x^2 x 1} dx]

Breaking this into two integrals:

[int frac{2x 1 3}{x^2 x 1} dx int frac{2x 1}{x^2 x 1} dx int frac{3}{x^2 x 1} dx]

The first integral is a standard logarithm form:

[int frac{2x 1}{x^2 x 1} dx ln |x^2 x 1| C_2]

The second integral requires a substitution:

[int frac{3}{x^2 x 1} dx 3 int frac{1}{left(x frac{1}{2}right)^2 left(frac{sqrt{3}}{2}right)^2} dx]

Using the formula for the inverse tangent function:

[int frac{1}{u^2 a^2} du frac{1}{a} tan^{-1} left(frac{u}{a}right) C]

We have:

[int frac{3}{left(x frac{1}{2}right)^2 left(frac{sqrt{3}}{2}right)^2} dx sqrt{3} tan^{-1} left(frac{2x 1}{sqrt{3}}right) C_3]

Combining all these results, we get:

[int frac{3x^7 dx}{x^3 - 1} frac{1}{3} left[ 7 ln |x - 1| - 7 ln |x^2 x 1| sqrt{3} tan^{-1} left(frac{2x 1}{sqrt{3}}right) right] C]

Thus, the final answer is:

[int frac{3x^7 dx}{x^3 - 1} frac{7}{3} ln |x - 1| - frac{7}{3} ln |x^2 x 1| frac{sqrt{3}}{3} tan^{-1} left(frac{2x 1}{sqrt{3}}right) C]

Conclusion

This detailed solution demonstrates the power of partial fractions in evaluating complex integrals. The integration process involves a series of algebraic manipulations and the application of standard integral forms. Understanding these techniques is crucial for advanced calculus and many real-world applications in science and engineering.