Chapter 5.3, Problem 26E

To determine

To evaluate : the integral

Answer to Problem 26E

The value of ${\displaystyle \underset{1}{\overset{2}{\int }}\frac{{\left(x-1\right)}^3}{x^2}}dx$ is $\ln 2-2$

Explanation of Solution

Given information : The integral is ${\displaystyle \underset{1}{\overset{2}{\int }}\frac{{\left(x-1\right)}^3}{x^2}}dx$

Calculation :

  $\begin{array}{l}{\displaystyle \underset{1}{\overset{2}{\int }}\frac{{\left(x-1\right)}^3}{x^2}}dx\\ \Rightarrow {\displaystyle \underset{1}{\overset{2}{\int }}\frac{\left(x-1\right)\left(x^2-2x-1\right)}{x^2}}dx\\ \Rightarrow {\displaystyle \underset{1}{\overset{2}{\int }}\frac{\left(x^3-2x^2-x-x^2+2x+1\right)}{x^2}}dx\\ \Rightarrow {\displaystyle \underset{1}{\overset{2}{\int }}\frac{\left(x^3-3x^2+x+1\right)}{x^2}}dx\\ \Rightarrow {\displaystyle \underset{1}{\overset{2}{\int }}\left[\frac{x^3}{x^2}-\frac{3x^2}{x^2}+\frac{x}{x^2}+\frac{1}{x^2}\right]}dx\\ \Rightarrow {\left[x-3+\frac{1}{x}+\frac{1}{x^2}\right]}_1^2\\ \Rightarrow {\left[\frac{1}{2}{x}^2-3x+\ln x+\frac{1}{x}\right]}_1^2\\ \Rightarrow 2-6+\ln 2+\frac{1}{2}-\left(\frac{1}{2}-3+\ln 1+1\right)\\ \Rightarrow \ln 2-2\end{array}$