Chapter G, Problem 31E

To determine

To calculate:

The integral value of the following:

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

Answer to Problem 31E

The integral value of ${\displaystyle \int \frac{dx}{x{(x^2+4)}^2}}$ is $\frac{\ln \left(\left|x\right|\right)}{16}-\frac{\ln \left(x^2+4\right)}{32}+\frac{1}{8x^2+32}+C$ .

Explanation of Solution

Given information:

The integral function:

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

Calculation:

The integral function is given that:

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

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

By this function, need to perform partial fraction decomposition:

  ${\displaystyle \int \frac{1}{x{(x^2+4)}^2}}dx={\displaystyle \int \left(-\frac{x}{16(x^2+4)}-\frac{x}{4{\left(x^2+4\right)}^2}+\frac{1}{16x}\right)}dx$

Apply the linearity:

  ${\displaystyle \int \frac{1}{x{(x^2+4)}^2}}dx=-\frac{1}{16}{\displaystyle \int \frac{x}{x^2+4}dx}-\frac{1}{4}{\displaystyle \int \frac{x}{{\left(x^2+4\right)}^2}}dx+\frac{1}{16}{\displaystyle \int \frac{1}{x}dx}$   ...(1)

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

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

The denominator value of above equation is rewritten as:

  $u=x^2+4$

Take the derivative:

  $\begin{array}{l}\frac{d}{dx}\left(u\right)=\frac{d}{dx}{x}^2+4\\ \frac{du}{dx}=2x\\ x=\frac{1}{2x}du\\ =\frac{1}{2}{\displaystyle \int \frac{1}{u}du}\end{array}$

Know that:

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

Plug in the solved integrals:

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

Put the value of $u$ :

  $u=x^2+4$

  $\frac{1}{2}{\displaystyle \int \text{}\frac{1}{u}du}=\frac{\ln \left(x^2+4\right)}{2}$

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

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

The denominator value of above equation is rewritten as:

  $u=x^2+4$

Take the derivative:

  $\begin{array}{l}\frac{d}{dx}\left(u\right)=\frac{d}{dx}{x}^2+4\\ \frac{du}{dx}=2x\\ x=\frac{1}{2x}du\\ =\frac{1}{2}{\displaystyle \int \frac{1}{u^2}du}\end{array}$

Know that:

  ${\displaystyle \int \frac{1}{u^2}du=-\frac{1}{u}}\left[\because {\displaystyle \int u^{n}du}=\frac{u^{n+1}}{n+1},n=-2\right]$

Plug in the solved integrals:

  $\frac{1}{2}{\displaystyle \int \frac{1}{u^2}du=-{\displaystyle \int \frac{1}{2u}}}$

Put the value of $u$ :

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

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

  ${\displaystyle \int \frac{1}{x}dx}$  ...(4)

Know that:

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

Put all the values in equation (1)

  $-\frac{1}{16}{\displaystyle \int \frac{x}{x^2+4}dx}-\frac{1}{4}{\displaystyle \int \frac{x}{{\left(x^2+4\right)}^2}dx+\frac{1}{16}{\displaystyle \int \frac{1}{x}dx}}=-\frac{\ln \left(x^2+4\right)}{32}+\frac{\ln \left(x\right)}{16}+\frac{1}{8\left(x^2+4\right)}$

This problem is solved. Apply the absolute value functions to logarithm functions arguments to extend the domain of anti-derivative:

  ${\displaystyle \int \frac{1}{x{\left(x^2+4\right)}^2}dx}=\frac{\ln \left(\left|x\right|\right)}{16}-\frac{\ln \left(x^2+4\right)}{32}+\frac{1}{8\left(x^2+4\right)}+C$

Simplify the equation:

  ${\displaystyle \int \frac{1}{x{\left(x^2+4\right)}^2}dx}=\frac{\ln \left(\left|x\right|\right)}{16}-\frac{\ln \left(x^2+4\right)}{32}+\frac{1}{8x^2+32}+C$