Chapter G, Problem 23E

To determine

To calculate:

Integral of following function.

  ${\displaystyle \int \frac{10}{\left(x-1\right)\left(x^2+9\right)}dx}$ .

Answer to Problem 23E

Integral of given function is $\frac{\ln \left(x^2+9\right)}{2}+\frac{\arctan \left(\frac{x}{3}\right)}{3}+\ln \left(\left|x-1\right|\right)+C$ .

Explanation of Solution

Given information:

  ${\displaystyle \int \frac{10}{\left(x-1\right)\left(x^2+9\right)}dx}$

Calculation:

Here given function is,

  ${\displaystyle \int \frac{10}{\left(x-1\right)\left(x^2+9\right)}dx}$

Apply integrity:

  $10{\displaystyle \int \frac{1}{\left(x-1\right)\left(x^2+9\right)}dx}$

After solving,

  ${\displaystyle \int \frac{1}{\left(x-1\right)\left(x^2+9\right)}dx}$

Partial fraction decomposition,

  $={\displaystyle \int \left(\frac{1}{10\left(x-1\right)}-\frac{x+1}{10\left(x^2+9\right)}\right)dx}$

Now applying linearity,

  $=\frac{1}{10}{\displaystyle \int \frac{1}{\left(x-1\right)}dx}-\frac{1}{10}{\displaystyle \int \frac{x+1}{\left(x^2+9\right)}dx}$

Now solving first term,

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

Substitute $u=x-1\to \frac{du}{dx}=1\to dx=du:$

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

Standard integral is,

  $=\ln \left(u\right)$

Undo substitution $u=x-1$ ,

  $=\ln \left(x-1\right)$

Now solving second term,

  ${\displaystyle \int \frac{x+1}{x^2+9}}dx$

Expand,

  $={\displaystyle \int \left(\frac{x}{x^2+9}+\frac{1}{x^2+9}\right)dx}$

Apply linearity,

  $={\displaystyle \int \frac{x}{x^2+9}dx+{\displaystyle \int \frac{1}{x^2+9}}dx}$

Now solving third term,

  ${\displaystyle \int \frac{x}{x^2+9}}dx$

Substitute the

  $\begin{array}{l}u=x^2+9+\to \frac{du}{dx}=2x\to dx=\frac{1}{2x}du:\\ =\frac{1}{2}{\displaystyle \int \frac{1}{u}du}\end{array}$

After solving,

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

Make use of previous result,

  $=\ln \left(u\right)$

Now plug all of the solved integrals,

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

  $\begin{array}{l}u=x^2+9\\ =\frac{\ln \left(x^2+9\right)}{2}dx\end{array}$

solving,

  ${\displaystyle \int \frac{x}{x^2+9}}dx$

Substitute,

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

Now solving,

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

This is standard integral,

  $=\arctan \left(u\right)$

Plug solve integral terms,

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

Undo substitution $u=\frac{x}{3}$ and plug solved integrals for first term,

  $\begin{array}{l}{\displaystyle \int \frac{x}{x^2+9}dx+{\displaystyle \int \frac{1}{x^2+9}dx}}\\ =\frac{\ln \left(x^2+9\right)}{2}+\frac{\arctan \left(\frac{x}{3}\right)}{3}\end{array}$

Plug solved integrals for second term,

  $\begin{array}{l}\frac{1}{10}{\displaystyle \int \frac{x}{x-1}dx-\frac{1}{10}{\displaystyle \int \frac{x+1}{x^2+9}dx}}\\ =\frac{\ln \left(x^2+9\right)}{2}+\frac{\arctan \left(\frac{x}{3}\right)}{3}+\frac{\ln \left(z-1\right)}{10}\end{array}$

Plug solved integrals for third term,

  $\begin{array}{l}10{\displaystyle \int \frac{1}{\left(x-1\right)\left(x^2+9\right)}dx}\\ =\frac{\ln \left(x^2+9\right)}{2}+\frac{\arctan \left(\frac{x}{3}\right)}{3}+\ln \left(x-1\right)\end{array}$

So integral for given function is:

  $\frac{\ln \left(x^2+9\right)}{2}+\frac{\arctan \left(\frac{x}{3}\right)}{3}+\ln \left(\left|x-1\right|\right)+C$