# Evaluate the following integral \int\limits_0^1 \frac{16(x-1)}{x^4-2x^3+4x-4} \text{d} x .

## Question:

Evaluate the following integral

{eq}\int\limits_0^1 \frac{16(x-1)}{x^4-2x^3+4x-4} \text{d} x {/eq}.

## Definite Integral

A definite integral of the form {eq}I=\int\limits_{x=a}^b f(x)~dx {/eq} where {eq}f(x) {/eq} is a rational function, that is, {eq}f(x)=\frac{p(x)}{q(x)} {/eq} and {eq}p(x) {/eq} and {eq}q(x) {/eq} are polynomials, can be evaluated by a number of techniques. One of the most common techniques is partial fractions. If the denominator, namely, {eq}q(x) {/eq} can be factored into one or more factors then the original integrand can be expressed as a sum of simpler fractions, each having a denominator that is one of the factors of {eq}q(x) {/eq}. After expressing the original integrand as a sum of partial fractions, the original integral can be expressed as a sum of simpler integrals. The remaining integrals can then be evaluated using other integration methods including substitution, integration by parts and the use of a table of integrals. These concepts are illustrated in the answer that follows.

The question is restated with slightly different notation. Evaluate the following definite integral

{eq}\begin{eqnarray*} I &=& \int_0^1\frac{16(x-1)}{x^4-2x^3+4x-4}dx. \end{eqnarray*} {/eq}

Consider the denominator of the integrand

{eq}\begin{eqnarray*} x^4-2x^3+4x-4 &=& x^4-2x^3+2x^2-2x^2+4x-4 \\ &=& x^2(x^2-2x+2)-2(x^2-2x+2) \\ &=& (x^2-2)(x^2-2x+2). \end{eqnarray*} {/eq}

Consequently, the integral can be written as

{eq}\begin{eqnarray*} I &=& \int_0^1\frac{16(x-1)}{(x^2-2)(x^2-2x+2)}dx \\ &=& \int_0^1\frac{16x-16}{(x^2-2)(x^2-2x+2)}dx. \end{eqnarray*} {/eq}

Now use partial fractions to write the integrand as

{eq}\begin{eqnarray*} \frac{16x-16}{(x^2-2)(x^2-2x+2)} &=& \frac{Ax}{x^2-2}+\frac{Bx+C}{x^2-2x+2} \\ \frac{16x-16}{(x^2-2)(x^2-2x+2)} &=& \frac{Ax(x^2-2x+2)+(Bx+C)(x^2-2)}{(x^2-2)(x^2-2x+2)} \\ 16x-16 &=& Ax(x^2-2x+2)+(Bx+C)(x^2-2) \\ 16x-16 &=& Ax^3-2Ax^2+2Ax+Bx^3-2Bx+Cx^2 \\ 16x-16 &=& (A+B)x^3+(C-2A)x^2+2(A-B)x-2C. \end{eqnarray*} {/eq}

The last equation yields

{eq}\begin{eqnarray*} A+B &=& 0 \\ C-2A &=& 0 \\ 2(A-B) &=& 16 \\ -2C &=& -16. \end{eqnarray*} {/eq}

From these equations it is easy to see that {eq}A=4 {/eq}, {eq}B=-4 {/eq} and {eq}C=8 {/eq}. Consequently

{eq}\begin{eqnarray*} I &=& \int_0^1\left(\frac{4x}{x^2-2}+\frac{-4x+8}{x^2-2x+2}\right)dx \\ &=& \int_0^1\frac{4x}{x^2-2}dx+\int_0^1\frac{-4x+8}{x^2-2x+2}dx \\ &=& 2\int_0^1\frac{2x}{x^2-2}dx+\int_0^1\frac{-4x+8}{x^2-2x+2}dx \\ &=& 2\left(\left.\ln|x^2-2|\right|_0^1\right)+\int_0^1\frac{-4x+8}{x^2-2x+2}dx \\ &=& -2\ln(2)+\int_0^1\frac{-4x+8}{x^2-2x+2}dx \\ &=& -2\ln(2)+8\int_0^1\frac{1}{x^2-2x+2}dx-4\int_0^1\frac{x}{x^2-2x+2}dx \\ &=& -2\ln(2)+8I_1-4I_2 \end{eqnarray*} {/eq}

where

{eq}\begin{eqnarray*} I_1 &=& \int_0^1\frac{1}{x^2-2x+2}dx \\ I_2 &=& \int_0^1\frac{x}{x^2-2x+2}dx. \end{eqnarray*} {/eq}

The inetgrals {eq}I_1 {/eq} and {eq}I_2 {/eq} can be evaluated using a table of integrals. Consider {eq}I_2 {/eq}. From a table we get

{eq}\begin{eqnarray*} I_2 &=& \int_{x=0}^1\frac{x}{x^2-2x+2}dx \\ &=& \frac{1}{2}\left(\left.\ln(x^2-2x+2)\right|_{x=0}^1\right)+I_1 \\ &=& -\frac{1}{2}\left(\ln(2)\right)+I_1 \\ \end{eqnarray*} {/eq}

Consider {eq}I_1 {/eq}. From a table we get

{eq}\begin{eqnarray*} I_1 &=& \int_{x=0}^1\frac{1}{x^2-2x+2}dx \\ &=& \frac{2}{\sqrt{4(1)(2)-(-2)^2}}\left(\tan^{-1}\left(\frac{2x-2}{\sqrt{4(1)(2)-(-2)^2}}\right)\right) \\ &=& \frac{2}{\sqrt{4}}\left(\tan^{-1}\left(\frac{2x-2}{\sqrt{4}}\right)\right) \\ &=& \left.\tan^{-1}(x-1)\right|_{x=0}^1 \\ &=& \tan^{-1}(0)-\tan^{-1}(-1) \\ &=& 0-\left(-\frac{\pi}{4}\right) \\ &=& \frac{\pi}{4}. \end{eqnarray*} {/eq}

Therefore

{eq}\begin{eqnarray*} I &=& -2\ln(2)+8I_1-4I_2 \\ &=& -2\ln(2)+8I_1-4\left(-\frac{1}{2}\left(\ln(2)\right)+I_1\right) \\ &=& -2\ln(2)+8I_1+2\ln(2)-4I_1 \\ &=& 8I_1-4I_1 \\ &=& 4I_1 \\ &=& 4\left(\frac{\pi}{4}\right) \\ &=& \pi. \end{eqnarray*} {/eq}