Does this telescope?

$l_{n}=\displaystyle\int_{0}^{1} \frac{x^{n}}{\sqrt{x^{3}+1}}dx$

Let us write $l_{n}$ as folllows:

$l_{n}=\displaystyle\int_{0}^{1} x^{n-2}\frac{x^{2}}{\sqrt{x^{3}+1}}dx$

Now using integration by parts:

$l_{n}=x^{n-2}\times\left.\displaystyle\frac{2\sqrt{x^{3}+1}}{3}\right|_{0}^{1}-\displaystyle\int_{0}^{1} (n-2)x^{n-3}\times\displaystyle\frac{2\sqrt{x^{3}+1}}{3}dx$

$\therefore l_{n}=\displaystyle\frac{2\sqrt{2}}{3}-\displaystyle\frac{2(n-2)}{3}\displaystyle\int_{0}^{1} \frac{x^{n-3}(x^{3}+1)}{\sqrt{x^{3}+1}}dx$

$\therefore l_{n}=\displaystyle\frac{2\sqrt{2}}{3}-\displaystyle\frac{2(n-2)}{3}\displaystyle\int_{0}^{1} \frac{x^{n-3}+x^{n}}{\sqrt{x^{3}+1}}dx$

$\therefore l_{n}=\displaystyle\frac{2\sqrt{2}}{3}-\displaystyle\frac{2(n-2)}{3}\left( l_{n-3}+l_{n}\right)$

Rearranging, we obtain:

$(2n-1)l_{n}+2(n-2)l_{n-3}=2\sqrt{2}$

Hence $(2m-1)l_{m}+2(m-2)l_{m-3}=\boxed{2\sqrt{2}}$