# Evaluate the integral \int_0^1 \frac {5-x}{1+x}^2 dx

## Question:

Evaluate the integral {eq}\int_0^1 \frac {5-x}{1+x}^2 dx {/eq}

## The Method of Simple Fractions:

The method of simple fractions to solve an integral is used when the fraction we want to integrate is proper. When the fraction is improper before using the method we must make the division first.

Transform the integrand

{eq}\displaystyle H=\frac {5-x}{1+x}^2\\ \displaystyle H=\frac {(5-x)^2}{(1+x)^2}\\ \displaystyle H=\frac {x^2-10x+25}{x^2+2x+1}\\ \displaystyle H=\frac {x^2-10x+25}{x^2+2x+1}\\ \displaystyle H=1+ \frac{-12x+24}{x^2+2x+1} {/eq}

Resolve into partial fractions.

{eq}\displaystyle \frac{-12x+24}{x^2+2x+1}= \frac {A}{x+1} +\frac {B}{(x+1)^2}\\ \displaystyle \frac{-12x+24}{x^2+2x+1}= \frac {A(x-1)+B}{ x^2+2x+1}\\ \displaystyle \frac{-12x+24}{x^2+2x+1}= \frac {Ax+A+B}{ x^2(x-1)}\\ \displaystyle A=-12\\ \displaystyle A+B=24 \, \Longrightarrow \, B=36\\ {/eq}

The fraction {eq}\frac{-12x+24}{x^2+2x+1} {/eq} into partial fractions are:

{eq}\displaystyle \Longrightarrow \boxed{ \frac {-12}{x+1} +\frac {36}{(x+1)^2}}\\ {/eq}

Calculate the integral

{eq}\displaystyle I = \int_0^1 \frac {5-x}{1+x}^2 \, dx\\ \displaystyle I = \int_0^1 1+ \frac {-12}{x+1} + \frac {36}{(x+1)^2} \, dx\\ \displaystyle I = \left. x - 12\ln |x+1| - \frac {72}{(x+1)^3} \right|_{0}^{1}\\ \displaystyle I = 1-12\ln |2| - 72 -(-9) \\ \displaystyle I = -12\ln |2| - 62 {/eq}

The result of the integral applying partial fraction method is:

{eq}\displaystyle \Longrightarrow \boxed{ I = -12\ln |2| - 62}\\ {/eq} 