Chapter G, Problem 26E

To determine

To calculate:

Integral of following function.

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

Answer to Problem 26E

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

Explanation of Solution

Given information:

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

Calculation:

Here given function is,

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

Partial fraction decomposition is,

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

Apply integrity:

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

After solving,

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

Thus us a standard integral,

  $=\arctan \left(x\right)$

Now solving,

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

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

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

Applying power rule,

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

with $n=-2$ ,

  $=-\frac{1}{u}$

Plug in solved integrals,

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

Undo substitution $u=x^2+1$ ,

  $=-\frac{1}{2\left(x^2+1\right)}$

Plug solved integrals in second term,

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