Question

If $I=\int (tanx{)}^{1/3}dx=A\log \frac{t^4-t^2+1}{{(t^2+1)}^2}+Bta{n}^{-1}\frac{2t^2-1}{2\sqrt{3}}+C$ where $t=ta{n}^{1/3}x$ then

• A. $A=1/4$
• B. $B=\sqrt{3}/2$
• C. $A=1/2$
• D. $B=2\sqrt{3}$

Answer 1

Answer: (A), (B)

$B=\sqrt{3}/2$
$A=1/4$

Let $I=\int {\left(\tan x\right)}^{\frac{1}{3}}dx$

Substitute $\tan x=t^3\Rightarrow 3t^{2}dt={\sec }^{2}xdx$

$I=3\int \frac{t^3}{1+t^6}dt$

Substitute $y=t^2$

$I=\frac{1}{2}\int \frac{y}{1+y^3}dy=\frac{1}{2}\int \frac{y+1}{y^2-y+1}dy-\frac{1}{2}\int \frac{dy}{y+1}$
$I=\frac{1}{4}\log \frac{t^4-t^2+1}{{(t^2+1)}^2}+\frac{\sqrt{3}}{2}{\tan }^{-1}\frac{2t^2-1}{2\sqrt{3}}$

Comparing with $I=\int (tanx{)}^{1/3}dx=A\log \frac{t^4-t^2+1}{{(t^2+1)}^2}+Bta{n}^{-1}\frac{2t^2-1}{2\sqrt{3}}+C$

$A=\frac{1}{4}$ and $B=\frac{\sqrt{3}}{2}$