Chapter G, Problem 22E

To determine

To calculate:

Integral of following function.

  ${\displaystyle \int \frac{x^2-x+6}{x^3+3x}dx}$ .

Answer to Problem 22E

Integral of given function is $2\ln \left(\left|x\right|\right)-\frac{\ln \left(x^2+3\right)}{2}-\frac{\arctan \left(\frac{x}{\sqrt{3}}\right)}{\sqrt{3}}+C$ .

Explanation of Solution

Given information:

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

Calculation:

Here given function is,

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

Now factor the denominator,

  $={\displaystyle \int \frac{x^2-x+6}{x\left(x^2+3\right)}dx}$

Partial fraction decomposition,

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

Now applying linearity,

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

Now solving first term,

  $={\displaystyle \int \frac{1}{x}dx}$

Standard integral is,

  $=\ln \left(x\right)$

Now solving second term,

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

Expand,

  $={\displaystyle \int \left(\frac{x}{x^2+3}+\frac{1}{x^2+3}\right)dx}$

Apply linearity,

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

Now solving third term,

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

Substitute the

  $\begin{array}{l}u=x^2+3+\to \frac{du}{dx}=2x\to dx=\frac{1}{2u}du:\\ =\frac{1}{2}{\displaystyle \int \frac{1}{u}du}\end{array}$

After solving,

  $={\displaystyle \int \frac{1}{u}du}$

Make use of previous result,

  $=\ln \left(u\right)$

Now plug all of the solved integrals,

  $=\frac{1}{2}{\displaystyle \int \frac{1}{u}dx}$

Substitute the $u=\frac{x}{\sqrt{3}}\to \frac{du}{dx}=\frac{1}{\sqrt{3}}\to dx=\sqrt{3}dx$

  $\begin{array}{l}={\displaystyle \int \frac{\sqrt{3}}{3u^2+3}du}\\ =\frac{1}{\sqrt{3}}{\displaystyle \int \frac{1}{u^2+1}}du\\ ={\displaystyle \int \frac{1}{u^2+1}}du\end{array}$

This is standard integrals,

  $=\arctan \left(u\right)$

Plug in all solved integrals

  $\begin{array}{l}\frac{1}{\sqrt{3}}{\displaystyle \int \frac{1}{u^2+1}du}\\ =\frac{\arctan \left(u\right)}{\sqrt{3}}\end{array}$

Undo substitution $u=\frac{x}{\sqrt{3}}$ ,

  $=\frac{\arctan \left(\frac{x}{\sqrt{3}}\right)}{\sqrt{3}}$

Plug solved integrals,

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