# Use partial fractions to solve. \int \frac{\sec^2 x}{\tan^3 x - \tan^2 x} dx show that by...

## Question:

Use partial fractions to solve.

{eq}\int \frac{\sec^2 x}{\tan^3 x - \tan^2 x} dx {/eq}

show that by substitution, {eq}\int \frac{\sec^{2x}}{\tan^{3x} - \tan^{2x}} dx = \int (\frac{A}{u} + \frac{B}{u^2} + \frac{C}{u-1}) du. {/eq}

Then find the solution by reversing the substitution in the final answer

A = B = -1 and C = 1.

## Partial Fraction Decomposition

Partial fraction decomposition, also called partial fraction expansion, is a technique used to rewrite a proper rational function as a sum of simpler functions. Essentially, it is the reverse procedure of finding a common denominator and adding fractions. To find the partial fraction decomposition of a proper rational function (a ratio of polynomials in which the degree of the numerator is smaller than the degree of the denominator), begin by factoring the denominator completely. If you have a linear factor {eq}ax+b {/eq}, this corresponds to a term in the expansion {eq}\frac{A}{ax+b} {/eq}. If the linear factor occurs multiple times, there is a corresponding term for each power of the factor. For instance, if you have {eq}(ax+b)^n {/eq} in your factored denominator, you have n terms {eq}\frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} + \ldots + \frac{A_n}{(ax+b)^n} {/eq} in the expansion. The procedure is similar for quadratic factors. An irreducible quadratic factor {eq}ax^2+bx+c {/eq} in the denominator will correspond to a term {eq}\frac{Ax+B}{ax^2+bx+c} {/eq} in the expansion - and there is one such term for each power of this factor.

{eq}\int \frac{\sec^2 x}{\tan^3 x - \tan^2 x} dx {/eq}

Let {eq}u=\tan(x)\\ du=\sec^2(x)dx {/eq}

Then we have that {eq}\int \frac{\sec^2 x}{\tan^3 x - \tan^2 x} dx = \int\frac{1}{u^3-u^2}du = \int\frac{1}{u^2(u-1)}du {/eq}

Finding the partial fraction decomposition of the integrand, we have

{eq}\frac{1}{u^2(u-1)} = \frac{A}{u} + \frac{B}{u^2} + \frac{C}{u-1}\\ 1=Au(u-1) + B(u-1) + Cu^2 {/eq}

Using the value {eq}u=0 {/eq} we have {eq}B= -1. {/eq} Using the value {eq}u=1 {/eq} we have {eq}C=1 {/eq}

Substituting these values in, distributing and comparing coefficients, we have {eq}1 = Au^2-Au-u+1+u^2\\ A=-1 {/eq}

Rewriting the integral, we have {eq}\int\frac{1}{u^2(u-1)}du = \int \left(\frac{A}{u}+\frac{B}{u^2}+\frac{C}{u-1}\right) du = \int \left(\frac{-1}{u}-\frac{1}{u^2}+\frac{1}{u-1}\right) du = -\ln|u|+\frac{1}{u}+\ln|u-1|+c = -\ln|\tan(x)| +\cot(x) +\ln|\tan(x)-1| +c {/eq} where the last equality comes from substituting our u value back into the answer.