Calculate integral using the value of another one

Question

Calculate the integral $\int_{-1}^{3} t^3(4 + t^3)^{-1/2} dt$, given that $\int_{-1}^{3} (4 + t^3)^{1/2} dt = 11.35$.

Leave the result in function of $√3 \text{ and }√31$.

I have tried $\int_{-1}^{3} (4 + t^3)^{1/2} dt = 11.35.$

$\int_{-1}^{3} \frac{(4 + t^3)} {(4 + t^3)^{1/2}} = 11.35.$

$\int_{-1}^{3} \frac{(4)} {(4 + t^3)^{1/2}} + \frac{t^{3}}{(4 + t^3)^{1/2}}= 11.35 .$

But I don't know how to continue. I would like to know if someone could recommend something to continue

Answer 1

$$\int_{－1}^{3}\frac {t^3}{\sqrt {4＋t^3}}dt＝\frac {1}{3}\int_{－1}^{3}t\frac {3t^2}{\sqrt {4＋t^3}}dt＝\frac {2}{3}\left({t\sqrt {4＋t^3}}\right)^3_{-1}-\frac {2}{3}\int_{-1}^{3}\sqrt{4＋t^3} \;dt$$