The statement says:using the calculus fundamental theorem find:

$\displaystyle \displaystyle\frac{d}{dx}(\displaystyle\int_{2}^{x }\displaystyle\frac{t^{3/2}}{\sqrt[ ]{t^2+17}}dt)$

I thought that what I should do is just to apply Barrow, and then I'd have:

$\displaystyle \displaystyle\frac{d}{dx}(\displaystyle\int_{2}^{x }\displaystyle\frac{t^{3/2}}{\sqrt[ ]{t^2+17}}dt)=-\displaystyle\frac{x^{3/2}}{\sqrt[ ]{x^2+17}}$

Is this right?

I've tried it on the hard way too, but then:

$\displaystyle t=\sqrt[ ]{17}\sinh u$

$\displaystyle dt=\sqrt[ ]{17}\cosh u du$

$\displaystyle \displaystyle\frac{d}{dx}(\displaystyle\int_{2}^{x }\displaystyle\frac{t^{3/2}}{\sqrt[ ]{t^2+17}dt})=\displaystyle\frac{d}{dx}(\sqrt[3]{17}\displaystyle\int_{2}^{x}\sqrt[3]{\sinh^2 u}du)$

I don't know how to solve the last integration.

Bye there.