Chapter G, Problem 34E

To determine

To calculate:

The integral value of the following:

  ${\displaystyle \int \frac{3x^2+x+4}{x^4+3x^2+2}dx}$

Answer to Problem 34E

The integral value of ${\displaystyle \int \frac{3x^2+x+4}{x^4+3x^2+2}dx}$ is $\frac{\ln \left(x^2+1\right)-\ln \left(x^2+2\right)}{2}+\sqrt{2}\arctan \left(\frac{x}{\sqrt{2}}\right)+\arctan \left(x\right)+C$ .

Explanation of Solution

Given information:

The integral function:

  ${\displaystyle \int \frac{3x^2+x+4}{x^4+3x^2+2}dx}$

Calculation:

The integral function is given that:

  ${\displaystyle \int \frac{3x^2+x+4}{x^4+3x^2+2}dx}$

Simply factor the denominator:

  ${\displaystyle \int \frac{3x^2+x+4}{x^4+3x^2+2}dx}={\displaystyle \int \frac{3x^2+x+4}{\left(x^2+1\right)\left(x^2+2\right)}dx}$

By this function, need to perform partial fraction decomposition:

  ${\displaystyle \int \frac{3x^2+x+4}{x^4+3x^2+2}dx}={\displaystyle \int \left(\frac{2-x}{x^2+2}+\frac{x+1}{x^2+1}\right)dx}$

Now, apply linearity:

  ${\displaystyle \int \frac{3x^2+x+4}{x^4+3x^2+2}dx}={\displaystyle \int \frac{x+1}{x^2+1}dx-}{\displaystyle \int \frac{x-2}{x^2+2}dx}$  ...(1)

Simplify the first term of above right hand side of equation (1)

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

Expand the equation:

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

Apply the linearity:

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

Simplify the first term of above right hand side of equation (2)

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

Substituting the value:

  $\begin{array}{l}u=x^2+1\\ \frac{d}{dx}\left(u\right)=\frac{d}{dx}\left(x^2+1\right)\\ \frac{du}{dx}=2x\\ dx=\frac{1}{2x}du\\ =\frac{1}{2}{\displaystyle \int \frac{1}{u}du}\end{array}$

Know that, this is the standard integral:

  ${\displaystyle \int \frac{1}{u}du}=\ln \left(u\right)$

Put the value:

  $\frac{1}{2}{\displaystyle \int \frac{1}{u}du=\frac{\ln \left(u\right)}{2}}$

Substitute the value $u=x^2+1$

  $\frac{1}{2}{\displaystyle \int \frac{1}{u}du=\frac{\ln \left(x^2+1\right)}{2}}$

Simplify the second term of above right hand side of equation (2)

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

Know that, this is the standard integral:

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

Put the value in equation (2)

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

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

Simplify the second term of above right hand side of equation (1)

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

Expand the equation:

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

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

Simplify the first term of above right hand side of equation (3)

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

Substituting the value:

  $\begin{array}{l}u=x^2+2\\ \frac{d}{dx}\left(u\right)=\frac{d}{dx}\left(x^2+2\right)\\ \frac{du}{dx}=2x\\ dx=\frac{1}{2x}du\\ =\frac{1}{2}{\displaystyle \int \frac{1}{u}du}\end{array}$

Use the previous results:

  $=\frac{\ln \left(u\right)}{2}$

Put the value $u=x^2+2$

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

Simplify the first term of above right hand side of equation (3)

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

Substitute the value:

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

Simplify the equation:

  ${\displaystyle \int \frac{1}{x^2+2}dx}=\frac{1}{\sqrt{2}}{\displaystyle \int \frac{1}{u^2+1}du}$   ...(4)

Know that,

  ${\displaystyle \int \frac{1}{u^2+1}du=\arctan \left(x\right)}$

Put the value:

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

Put the value in equation (3)

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

  ${\displaystyle \int \frac{x-2}{x^2+2}dx}=\frac{\ln \left(x^2+2\right)}{2}-\sqrt{2}\arctan \left(\frac{x}{\sqrt{2}}\right)$

Put this value in equation (1)

  ${\displaystyle \int \frac{3x^2+x+4}{x^4+3x^2+2}dx}={\displaystyle \int \frac{x+1}{x^2+1}dx-}{\displaystyle \int \frac{x-2}{x^2+2}dx}$

  ${\displaystyle \int \frac{3x^2+x+4}{x^4+3x^2+2}dx}=-\frac{\ln \left(x^2+2\right)}{2}+\frac{\ln \left(x^2+1\right)}{2}+\sqrt{2}\arctan \left(\frac{x}{\sqrt{2}}\right)+\arctan \left(x\right)$

  ${\displaystyle \int \frac{3x^2+x+4}{x^4+3x^2+2}dx}=-\frac{\ln \left(x^2+2\right)}{2}+\frac{\ln \left(x^2+1\right)}{2}+\sqrt{2}\arctan \left(\frac{x}{\sqrt{2}}\right)+\arctan \left(x\right)+C$

  ${\displaystyle \int \frac{3x^2+x+4}{x^4+3x^2+2}dx}=\frac{\ln \left(x^2+1\right)-\ln \left(x^2+2\right)}{2}+\sqrt{2}\arctan \left(\frac{x}{\sqrt{2}}\right)+\arctan \left(x\right)+C$