Question

Calculate each of the definite integrals in Exercises 47–52. Some integrals require partial fractions or polynomial long division, and some do not.

${\int }_{-1}^1\frac{4(x^2-1)}{x^2-4}dx$

Short answer

The value is$-\frac{3}{2}\ln \left(3\right)$

Step-by-step solution

Step 1. Given Information: 

Given definite integral :${\int }_{-1}^1\frac{4(x^2-1)}{x^2-4}dx$

We want to calculate each of the definite integrals.

Step 2. Calculation: 

We have

$$\begin{gathered} {\int }_{-1}^1\frac{4(x^2-1)}{x^2-4}dx=4{\int }_{-1}^1\frac{x^2-1}{x^2-4}dx \\ Uselongdivisionmethodweget \\ \frac{x^2-1}{x^2-4}=1+\frac{3}{x^2-4} \\ Now \\ 4{\int }_{-1}^1\frac{x^2-1}{x^2-4}dx=4{\int }_{-1}^1\left(1+\frac{3}{x^2-4}\right)dx \\ =4\left({\int }_{-1}^{1}1dx+{\int }_{-1}^1\frac{3}{x^2-4}dx\right) \\ Solve \\ {\int }_{-1}^1\frac{3}{x^2-4}dx=3{\int }_{-1}^1\frac{1}{x^2-4}dx \\ Substitutionx=2u \\ Whenx=1thenu=\frac{1}{2}andwhenx=-1thenu=-\frac{1}{2} \\ andalsodx=2du \\ ApplyIntegralSubstitution: \\ 3{\int }_{-1}^1\frac{1}{x^2-4}dx=3{\int }_{-\frac{1}{2}}^{\frac{1}{2}}\frac{1}{4u^2-4}2du \\ =\frac{3}{2}{\int }_{-\frac{1}{2}}^{\frac{1}{2}}\frac{1}{u^2-1}du \\ =-\frac{3}{2}{\int }_{-\frac{1}{2}}^{\frac{1}{2}}\frac{1}{1-u^2}du \\ =-\frac{3}{2}{\left[\frac{\ln \left|u+1\right|}{2}-\frac{\ln \left|u-1\right|}{2}\right]}_{-\frac{1}{2}}^{\frac{1}{2}} \\ Simplify: \\ =-\frac{3}{2}\left[\frac{\ln \left|{\displaystyle \frac{3}{2}}\right|}{2}-\frac{\ln \left|-{\displaystyle \frac{1}{2}}\right|}{2}-\left(\frac{\ln \left|{\displaystyle \frac{1}{2}}\right|}{2}-\frac{\ln \left|-{\displaystyle \frac{3}{2}}\right|}{2}\right)\right] \\ Use\log properties: \\ =-\frac{3}{2}\left(\frac{2\ln \left(2\right)+2\ln \left({\displaystyle \frac{3}{2}}\right)}{2}\right) \\ =-\frac{3}{2}\ln \left(3\right) \end{gathered}$$