Exercise 7.4 Integrate the function $\frac{3x}{1+2x^4}$

Given function is $\frac{3x}{1+2x^4}$

The integral of  $\frac{3x}{1+2x^4}$ is $\int \frac{3x}{1+2x^4}dx$

Step 1: Use substitution method to convert the integral into some of the known integral

$\int \frac{3x}{1+2x^4}dx=\int \frac{3x}{1+2(x^2{)}^2}dx$

$=\int \frac{3x}{1+(\sqrt{2}{x}^2{)}^2}dx$

Let $x^2=t$, then $\frac{d}{dx}{x}^2=\frac{dt}{dx}\Rightarrow xdx=\frac{dt}{2}$

Therefore, $\int \frac{3x}{1+2x^4}dx=\frac{3}{2}\int \frac{1}{1+(\sqrt{2}t{)}^2}dt$.  ....(1)

Step 2: Compare the integral in (1) with the known integral 

The integral $\int \frac{1}{1+(\sqrt{2}t{)}^2}dt$ is of the form $\int \frac{1}{1+(ax{)}^2}dx=\frac{1}{a}{\tan }^{-1}(ax)$

Therefore, $\int \frac{1}{1+(\sqrt{2}t{)}^2}dt=\frac{1}{\sqrt{2}}{\tan }^{-1}(\sqrt{2}t)$

Substitute back $t=x^2$, we get

$\int \frac{1}{1+(\sqrt{2}{x}^2{)}^2}dt=\frac{1}{\sqrt{2}}{\tan }^{-1}(\sqrt{2}{x}^2)+C$

$\Rightarrow \int \frac{3x}{1+2x^4}dx=\frac{3}{2}\times \frac{1}{\sqrt{2}}{\tan }^{-1}(\sqrt{2}{x}^2)+C$

Therefore, $\int \frac{3x}{1+2x^4}dx=\frac{3}{3\sqrt{2}}{\tan }^{-1}(\sqrt{2}{x}^2)+C$.