$$\int\_0^1 \frac{\sqrt{x}}{1+\sqrt[3]{x}}dx$$

I Know I should use this substitution: $x=t^6$, $dx=6t^5dt$ integral

$$\int\_{{x\_1}=0}^{{x\_2}=1} \frac{\sqrt{x}}{1+\sqrt[3]{x}}dx=\int\_{t\_1}^{t\_2} \frac{t^3(6t^5dt)}{1+t^2}$$

And also I can solve This integral

But I have problem with limits ( $t\_1$,$t\_2$) : $x\_1=0$ so ${t\_1}^6=0$ and $t\_1=0$ and for $x\_2=1$ so ${t\_2}^6=1$ and $t\_2=+1 $ and $t\_2=-1 $

here I have problem whcich one Shluld I use as integral limit $t\_2=+1 $ or $t\_2=-1 $ and Why?