Solving the Integral of ∫ dx/(x-1)(x-2)(x-3) Using Partial Fractions

When dealing with integrals of forms such as ∫ dx/(x-1)(x-2)(x-3), a common technique is to use partial fraction decomposition. This article will explain the method to solve this type of integral step-by-step.

Introduction to Partial Fractions

Partial fraction decomposition is a method used to simplify complex rational expressions. For an integral of the form ∫ dx/(x-1)(x-2)(x-3), we can break it down into simpler terms using the following structure:

displaystyle frac{1}{(x-1)(x-2)(x-3)} frac{A}{x-1} frac{B}{x-2} frac{C}{x-3}

Step-by-Step Calculation

First, we multiply both sides by the common denominator (x-1)(x-2)(x-3) to formulate the following equation:

1 A(x-2)(x-3) B(x-1)(x-3) C(x-1)(x-2)

We then find the values of A, B, and C by substituting specific values of x. Here are the steps:

Substitution for A

Let x 1 1 A(1-2)(1-3) 1 A(-1)(-2) 1 2A A 1/2

Substitution for B

Let x 2 1 B(2-1)(2-3) 1 B(1)(-1) 1 -B B -1

Substitution for C

Let x 3 1 C(3-1)(3-2) 1 C(2)(1) 1 2C C 1/2

Using these values for A, B, and C, we can rewrite the integral as:

The Integral

displaystyle int frac{1}{(x-1)(x-2)(x-3)} dx int frac{1/2}{x-1} - frac{1}{x-2} frac{1/2}{x-3} dx

This can be further simplified:

displaystyle (1/2)ln (x-1) - ln (x-2) (1/2)ln (x-3) C

Alternative Approach

Alternatively, an easier method to solve the integral involves using a substitution. Let us use the substitution u x - 2, so the expression becomes:

displaystyle int frac{1}{(u-1)(u)(u 1)} du

Using partial fractions, we can decompose this expression as:

displaystyle frac{1}{(u-1)(u)(u 1)} frac{1/2}{u-1} - frac{1}{u} frac{1/2}{u 1}

The integral then becomes:

displaystyle int [frac{1/2}{u-1} - frac{1}{u} frac{1/2}{u 1}] du

Separating the terms, we get:

displaystyle (1/2)ln (u-1) - ln (u) (1/2)ln (u 1) C

Substituting back u x - 2, we obtain the final result:

displaystyle (1/2)ln (x-3) - ln (x-2) (1/2)ln (x-1) C

Conclusion

The integral of ∫ dx/(x-1)(x-2)(x-3) can be solved using partial fractions or an alternative substitution method. Both methods show the same final result: