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.

$\int \frac{x^4-3}{2+3x^2}dx$

Short answer

$\frac{x^3}{9}-\frac{2}{9}x-\frac{23}{54}\sqrt{6}{\tan }^{-1}\left(\frac{\sqrt{6}}{2}x\right)+C$

Step-by-step solution

Step1. Given Information

The integral is as follows.

$\int \frac{x^4-3}{2+3x^2}dx$

The objective is to solve the integral.

Step2. Long division

The polynomial long division method is calculated below.

$\begin{array}{r}\frac{\frac{1}{3}{x}^2-\frac{2}{9}}{3x^2+2\overline{)x^4-3}}\\ \frac{x^4+\frac{2}{3}{x}^2}{-\frac{2}{3}{x}^2-3}\\ \frac{-\frac{2}{3}{x}^2-\frac{4}{9}}{-\frac{23}{9}}\end{array}$

The expression is in the form of$\frac{x^4-3}{2+3x^2}=\frac{x^2}{3}-\frac{2}{9}+\frac{-\frac{23}{9}}{3x^2+2}.$

Step3. Solution

The integral is solved below.

$\begin{array}{r}\int \frac{x^4-3}{2+3x^2}dx\\ =\int \left(\frac{x^2}{3}-\frac{2}{9}+\frac{-\frac{23}{9}}{3x^2+2}\right)dx\\ =\int \frac{x^2}{3}dx-\int \frac{2}{9}dx-\frac{23}{9}\int \frac{1}{3x^2+2}dx\\ =\frac{x^3}{9}-\frac{2}{9}x-\frac{23}{54}\sqrt{6}{\tan }^{-1}\left(\frac{\sqrt{6}}{2}x\right)+C\end{array}$

Therefore, the value is$\frac{x^3}{9}-\frac{2}{9}x-\frac{23}{54}\sqrt{6}{\tan }^{-1}\left(\frac{\sqrt{6}}{2}x\right)+C$