Question

Integrate w.r.t x $\frac{3x}{1+2x^4}$

• A. $\frac{3}{\sqrt{2}}ta{n}^{-1}\sqrt{2}{x}^2+c$
• B. $\frac{3}{2\sqrt{2}}ta{n}^{-1}\sqrt{2}{x}^2+c$
• C. $\frac{3}{2\sqrt{2}}ta{n}^{-1}2x^2+c$
• D. $\frac{3}{\sqrt{2}}ta{n}^{-1}{x}^2+c$

Answer 1

The correct option is B $\frac{3}{2\sqrt{2}}ta{n}^{-1}\sqrt{2}{x}^2+c$

Putting $\sqrt{2}{x}^2=t$

$⟹2\sqrt{2}xdx=dt$

$⟹\int \frac{3dt}{2\sqrt{2}(1+t^2)}$

$⟹\frac{3}{2\sqrt{2}}ta{n}^{-1}\sqrt{2}{x}^2$