Chapter 6, Problem 26RE

To determine

To calculate: Evaluate the integral ${\displaystyle \underset{1}{\overset{2}{\int }}\left(x+\frac{1}{x^2}\right)dx}$

Answer to Problem 26RE

  $2$

Explanation of Solution

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

Formula used: Use integration formula ${\displaystyle \int x^{n}dx}=\frac{x^{n+1}}{n+1}+c$

Calculation: Use property ${\displaystyle \int \left(f\left(x\right)+g\left(x\right)\right)dx}={\displaystyle \int f\left(x\right)dx}+{\displaystyle \int g\left(x\right)dx}$

  ${\displaystyle \underset{1}{\overset{2}{\int }}\left(x+\frac{1}{x^2}\right)dx}={\displaystyle \underset{1}{\overset{2}{\int }}xdx}+{\displaystyle \underset{1}{\overset{2}{\int }}\frac{1}{x^2}dx}$

Hence,

  $\begin{array}{c}{\displaystyle \underset{1}{\overset{2}{\int }}xdx}+{\displaystyle \underset{1}{\overset{2}{\int }}\frac{1}{x^2}dx}={\displaystyle \underset{1}{\overset{2}{\int }}xdx}+{\displaystyle \underset{1}{\overset{2}{\int }}{x}^{-2}dx}\\ ={\left[\frac{x^{1+1}}{1+1}\right]}_1^2+{\left[\frac{x^{-2+1}}{-2+1}\right]}_1^2\\ ={\left[\frac{x^2}{2}\right]}_1^2+{\left[\frac{x^{-1}}{-1}\right]}_1^2\\ ={\left[\frac{x^2}{2}\right]}_1^2-{\left[\frac{1}{x}\right]}_1^2\end{array}$

Further simplify it

  $\begin{array}{c}{\left[\frac{x^2}{2}\right]}_1^2-{\left[\frac{1}{x}\right]}_1^2=\left[\frac{2^2}{2}-\frac{1^2}{2}\right]-\left[\frac{1}{2}-\frac{1}{1}\right]\left(\text{Substitute the upper and lower limit}\right)\\ =\left[2-\frac{1}{2}\right]-\left[\frac{1}{2}-1\right]\\ =\frac{3}{2}-\left[-\frac{1}{2}\right]\\ =\frac{3}{2}+\frac{1}{2}\\ =2\end{array}$

Therefore, ${\displaystyle \underset{1}{\overset{2}{\int }}\left(x+\frac{1}{x^2}\right)dx}=2$