Integration of 1/(x^3(1-x)) using Partial Fractions and Alternative Methods

Introduction to Integration Technique

Integration is a fundamental component of calculus which helps in solving a wide range of problems in physics, engineering, and other fields. This article focuses on the integration of the function 1/(x^3(1-x)). We will explore the integration process through the method of partial fractions and an alternative method involving the manipulation of the numerator.

Method 1: By Manipulating the Numerator

Consider the integral displaystyleint dfrac{mathrm dx}{x^3 cdot 1 x}. We start with the manipulation of the numerator to express it in a form that simplifies the integration. Let's rewrite the numerator as x^3 - x^3.

[ I displaystyleint dfrac{x dx - x dx}{x^3 cdot 1 x} displaystyleint dfrac{x dx}{x^3 cdot 1 x} - displaystyleint dfrac{dx}{x^3 cdot 1 x} ]

By further simplification, we can integrate each term separately:

[ I displaystyleint dfrac{x^2 - x 1}{x^3} , dx - ln |1 - x| ]

[ I displaystyleint dfrac{1}{x} , dx - displaystyleint dfrac{1}{x^2} , dx - displaystyleint dfrac{1}{x^3} , dx - ln |1 - x| C ]

Performing the integrations, we get:

[ I ln |x| - dfrac{1}{x} dfrac{1}{2x^2} - ln |1 - x| C ]

Method 2: By Applying Partial Fractions

The second method involves decomposing the integral using partial fractions. We start by expressing the integrand as:

[ dfrac{1}{x^3(1-x)} dfrac{A}{x} dfrac{B}{x^2} dfrac{C}{x^3} dfrac{D}{1-x} ]

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

[ 1 Ax^2(1-x) Bx(1-x) C(1-x) Dx^3 ]

[ 1 Ax^2 - Ax^3 Bx - Bx^2 C - Cx Dx^3 ]

To find the constants A, B, C, D, we equate the coefficients of like terms:

[ 1 (A - B)x^2 (B - C)x (C - A D) ]

From the above, we equate the coefficients to zero (since there is no x^2 or x term on the right side):

( A - B 0 ) ( B - C 0 ) ( C - A D 1 )

We also use specific values of x to determine the constants:

When x -1 : ( D - 1 1 ) rarr; ( D -1 ) When x 0 : ( C - 0 1 ) rarr; ( C 1 ) When x 1 : ( A - 1 D 1 ) rarr; ( A - 1 - 1 1 ) rarr; ( A 1 ) ( B -1 )

Substituting these values back into the partial fraction, we get:

[ dfrac{1}{x^3(1-x)} dfrac{1}{x} - dfrac{1}{x^2} - dfrac{1}{x^3} - dfrac{1}{1-x} ]

Integrating each term, we have:

[ I ln |x| - dfrac{1}{x} dfrac{1}{2x^2} - ln |1 - x| C ]

Conclusion

Both methods yield the same simplified form for the indefinite integral. The integration of 1/(x^3(1-x)) can be carried out using either the method of manipulating the numerator or the method of partial fractions. Both methods provide a systematic approach to solving such integrals and can be adapted to integrate various complex functions in calculus.