Question

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.

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

Short answer

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

Step-by-step solution

Step1. Given Information

The integral is as follows.

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

The objective is ton solve the integral.

Step2. Long division

The polynomial long division method is calculated below.

$$\begin{gathered} x-3 \\ \begin{array}{c}{x}^2+1\overline{)x^3-3x^2+2x-3}\\ \frac{x^3+x}{-3x^2+x}\\ \frac{-3x^2-3}{x}\end{array} \end{gathered}$$

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

Step3. Solution

The integral is solved below.

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

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