Question

Integrate the following functions.
$\int \frac{3x}{1+2x^4}dx.$

Answer 1

$\int \frac{3x}{1+2x^4}dx=\frac{3}{2}\int \frac{xdx}{\frac{1}{2}+x^4}=\frac{3}{2}\int \frac{xdx}{\frac{1}{2}+(x^2{)}^2}Let{x}^2=t\Rightarrow 2x=\frac{dt}{dx}\Rightarrow dx=\frac{dt}{2x}\therefore \frac{3}{2}\int \frac{xdx}{\frac{1}{2}+(x^2{)}^2}=\frac{3}{2}\int \frac{x}{\frac{1}{2}+(t{)}^2}\frac{dt}{2x}=\frac{3}{4}\int \frac{dt}{t^2+(\frac{1}{\sqrt{2}}{)}^2}=\frac{3}{4}\times \frac{1}{\frac{1}{\sqrt{2}}}ta{n}^{-1}\left(\frac{1}{\frac{1}{\sqrt{2}}}\right)+C[\because \int \frac{dx}{a^2+x^2}=\frac{1}{a}ta{n}^{-1}ta{n}^{-1}(\frac{x}{a}\left)\right]=\frac{3}{2\sqrt{2}}ta{n}^{-1}\left(\sqrt{2}{x}^2\right)+C(\because t=x^2)$