Question

Evaluate $\int \frac{1+x^{-2/3}}{1+x}dx.$
The ans is $=\frac{1}{2}\left[log\right(t+1{)}^4(t^2-t+1)]+\sqrt{3}.{\tan }^{-1}\frac{2t-1}{\sqrt{\left(3\right))}}$,
then $t=?$

• A. $x^{1/3}$
• B. $x^{1/2}$
• C. $x$
• D. $x^3$

Answer 1

The correct option is A $x^{1/3}$

Let $I=\int \frac{1+x^{\frac{-2}{3}}}{1+x}dx.$

Substitute $x=t^3$

$\therefore I=3\int \frac{t^2+1}{(t^3+1)}dt=3\int \frac{(t^2+2)}{(t+1)(t^2-t+1)}$

Using partial fraction
$I=3\int (\frac{2}{3}\frac{1}{t+1}+\frac{1}{3}.\frac{t+1}{t^2-t+1})dt$

$=2\log (t+1)+\frac{1}{2}\int \frac{2t-1+3}{(t^2-t+1)}dt$

$=2\log (t+1)+\frac{1}{2}\log (t^2-t+1)+\frac{3}{2}\int \frac{dt}{{(t-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$

$=\frac{1}{2}\left[4\log (t+1)+\log (t^2-t+1)\right]+\frac{3}{2}.\frac{2}{\sqrt{3}}{\tan }^{-1}\frac{t-\frac{1}{2}}{\sqrt{\frac{3}{2}}}$

$=\frac{1}{2}\log \left[{(t+1)}^4(t^2-t+1)\right]+\sqrt{3}{\tan }^{-1}\frac{2t-1}{\sqrt{3}}$
Where $t=x^{\frac{1}{3}}$