Chapter 5.7, Problem 28E

To determine

To calculate: The value of given integral

Answer to Problem 28E

The value of the given integral is $2\ln \left|x\right|-\frac{1}{2}\ln \left|x^2+3\right|-\frac{1}{\text{}\sqrt{3}}{\tan }^{-1}\left(\frac{x}{\sqrt{3}}\right)+C$

Explanation of Solution

Given information:

The integral is ${\displaystyle \int \frac{x^2-x+6}{x^3+3x}dx}$

Formula used:

  ${\displaystyle \int \frac{1}{x}dx=\ln \left|x\right|+C}$

  ${\displaystyle \int \frac{dx}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)}$

Calculation:

Since $x^3+3x=x\left(x^2+3\right)$ can’t be factored further, so write

  $\frac{x^2-x+6}{x\left(x^2+3\right)}=\frac{A}{x}+\frac{Bx+C}{x^2+3}$

Multiply by $x\left(x^2+3\right)$ , so

  $\begin{array}{l}{x}^2-x+6=A\left(x^2+3\right)+\left(Bx+C\right)x\\ x^2-x+6=\left(A+B\right)x^2+Cx+3A\end{array}$

Now, equate the coefficients,

  $A+B=1,C=-1\text{ and }3A=6$

Thus, $A=2,B=-1\text{ and }C=-1$ and so,

  ${\displaystyle \int \frac{x^2-x+6}{x^3+3x}dx}={\displaystyle \int \left[\frac{2}{x}+\frac{-x-1}{x^2+3}\right]}dx$

In order to integrate the second term split it into two parts

  ${\displaystyle \int \frac{-x-1}{x^2+3}dx=-{\displaystyle \int \frac{x}{x^2+3}dx-{\displaystyle \int \frac{1}{x^2+3}dx}}}$

Make the substitution $u=x^2+3$ in the first of these integrals so that $du=2xdx$ . Evaluate the second integral by means of the formula

  $\begin{array}{l}{\displaystyle \int \frac{x^2-x+6}{x^3+3x}dx}\\ =2{\displaystyle \int \frac{1}{x}dx-{\displaystyle \int \frac{x}{x^2+3}dx-{\displaystyle \int \frac{1}{x^2+3}dx}}}\\ =2\ln \left|x\right|-\frac{1}{2}\ln \left|x^2+3\right|-\frac{1}{\text{}\sqrt{3}}{\tan }^{-1}\left(\frac{x}{\sqrt{3}}\right)+C\end{array}$

Therefore, the value of the given integral is $2\ln \left|x\right|-\frac{1}{2}\ln \left|x^2+3\right|-\frac{1}{\text{}\sqrt{3}}{\tan }^{-1}\left(\frac{x}{\sqrt{3}}\right)+C$