Evaluating the Integral Using Partial Fractions: A Step-by-Step Guide

In this article, we will delve into the process of evaluating an integral using the method of partial fractions. We'll tackle the specific example of the integral of -2x^4 / (x^2 1)(x - 1)^2 dx, breaking it down into comprehensible steps suitable for those just starting to explore more complex integration techniques in calculus.

Introduction to Partial Fractions

Partial fractions is a technique used in integral calculus to simplify the process of integration of rational functions. By expressing a rational function as a sum of simpler fractions, we can often solve the integral more easily. In this article, we'll apply this technique to the given integral and demonstrate the steps involved.

Problem Statement and Initial Setup

Let's consider the integral:

∫(-2x^4) / (x^2 1)(x - 1)^2 dx

We will use partial fractions to decompose the integrand:

-2x^4 / (x^2 1)(x - 1)^2 A / (x - 1) B / (x - 1)^2 (Cx D) / (x^2 1)

Solving for Coefficients A, B, C, and D

To solve for the constants A, B, C, and D, we equate the numerators:

-2x^4 A(x^2 1) B(x^2 1) (Cx D)(x - 1)^2

Step 1: Finding B

Set x 1 to find B:

-2(1)^4 B(1^2 1) > B 1

Step 2: Solving for A and C

Comparing coefficients of x^3, we get:

A C 0 > C -A

Comparing coefficients of x^2:

-A B - 2C D 0 > -A 1 - 2C D 0

Using B 1 and C -A:

-A 1 - 2(-A) D 0 > -A 1 2A D 0 > A D -1

Step 3: Solving for C and D

Comparing the coefficients of x:

-2 A - 2D > -A - 2D -2 (Equation 3)

Comparing the constant term:

4 -A D (Equation 4)

From Equation 4, we get: D 4 A

Substitute D 4 A into A D -1 to find A:

A (4 A) -1 > 2A 4 -1 > 2A -5 > A -2.5

Therefore, D 4 - 2.5 1.5

Using A -2.5:

-2.5 - 5 -7 > C -(-2.5) 2.5

Hence:

C 2, D 1.5

Final Partial Fraction Decomposition

Substituting the values found, we have:

-2x^4 / (x^2 1)(x - 1)^2 -2.5 / (x - 1) 1 / (x - 1)^2 (2x 1.5) / (x^2 1)