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