Chapter G, Problem 24E

To determine

To calculate:

Integral of following function.

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

Answer to Problem 24E

Integral of given function is $\ln \left(x^2+1\right)-\ln \left(\left|x-1\right|\right)+C$ .

Explanation of Solution

Given information:

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

Calculation:

Here given function is,

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

Partial fraction decomposition is,

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

Apply integrity:

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

After solving,

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

Substitute the $u=x^2+1\to \frac{du}{dx}=2x\to dx=\frac{1}{2x}du$

  $\begin{array}{l}=\frac{1}{2}{\displaystyle \int \frac{1}{u}dx}\\ ={\displaystyle \int \frac{1}{u}dx}\\ =\ln \left(u\right)\end{array}$

Now plug solved integrals in first term,

  $\begin{array}{l}=\frac{1}{2}{\displaystyle \int \frac{1}{u}dx}\\ =\frac{\ln \left(u\right)}{2}\end{array}$

Now undo the substitution $u=x^2+1:$

  $=\frac{\ln \left(x^2+1\right)}{2}$

Now solving,

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

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

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

Use the previous result:

  $=\ln \left(u\right)$

Undo the substitution $u=x-1$ ,

  $=\ln \left(x-1\right)$

Plug in solved integrals,

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