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Q. 78

Expert-verified
Found in: Page 465

### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

# Solve given integrals by using polynomial long division to rewrite the integrand. This is one way that you can sometimes avoid using trigonometric substitution; moreover, sometimes it works when trigonometric substitution does not apply. $\int \frac{{x}^{3}-3{x}^{2}+2x-3}{{x}^{2}+1}$dx

$\frac{{x}^{2}}{2}-3x+\frac{1}{2}\mathrm{ln}\left(\left|{x}^{2}+1\right|\right)+C\text{.}$

See the step by step solution

## Step1. Given Information

The integral is as follows.

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

The objective is ton solve the integral.

## Step2. Long division

The polynomial long division method is calculated below.

$x-3\phantom{\rule{0ex}{0ex}}\begin{array}{c}{x}^{2}+1\overline{){x}^{3}-3{x}^{2}+2x-3}\\ \frac{{x}^{3}+x}{-3{x}^{2}+x}\\ \frac{-3{x}^{2}-3}{x}\end{array}$

The expression is in the form of $\frac{{x}^{3}-3{x}^{2}+2x-3}{{x}^{2}+1}=x-3+\frac{x}{{x}^{2}+1}$

## Step3. Solution

The integral is solved below.

$\begin{array}{l}\int \frac{{x}^{3}-3{x}^{2}+2x-3}{{x}^{2}+1}dx\\ =\int \left(x-3+\frac{x}{{x}^{2}+1}\right)dx\\ =\int xdx-\int 3dx+\int \frac{x}{{x}^{2}+1}dx\\ =\frac{{x}^{2}}{2}-3x+\frac{1}{2}\int \frac{1}{u}du\phantom{\rule{1em}{0ex}}\left[u={x}^{2}+1,du=2xdx\right]\\ =\frac{{x}^{2}}{2}-3x+\frac{1}{2}\mathrm{ln}\left(|u|\right)+C\phantom{\rule{1em}{0ex}}\frac{{x}^{2}}{2}-3x+\frac{1}{2}\mathrm{ln}\left(\left|{x}^{2}+1\right|\right)+C\end{array}$

Therefore, the value is $\frac{{x}^{2}}{2}-3x+\frac{1}{2}\mathrm{ln}\left(\left|{x}^{2}+1\right|\right)+C\text{.}$