1. ## tough integral?.

Here's an integral to try. I couldn't even get Maple to give me a closed form for it. Kriz will probably conquer it like Grant through Richmond.

$\int\frac{\sqrt{1+x^{2}}}{\sqrt{10-x^{3}}}dx$

2. Originally Posted by galactus
Here's an integral to try. I couldn't even get Maple to give me a closed form for it. Kriz will probably conquer it like Grant through Richmond.

$\int\frac{\sqrt{1+x^{2}}}{\sqrt{10-x^{3}}}dx$
My guess would be to start thusly:

$\int\frac{\sqrt{1+x^2}}{\sqrt{10-x^3}}dx$

$\int\frac{\sqrt{1+x^2}}{\sqrt{10-x^3}}\frac{\sqrt{1+x^2}}{\sqrt{1+x^2}}
dx$

$\int\frac{1+x^2}{\sqrt{(10-x^3)(1+x^2)}}dx$

$\int\frac{1}{\sqrt{(10-x^3)(1+x^2)}}dx+\int\frac{x^2}{\sqrt{(10-x^3)(1+x^2)}}dx$

How to finish I cannot tell immediately, or even whether the above is an appropriate beginning.

3. Originally Posted by galactus
Here's an integral to try. I couldn't even get Maple to give me a closed form for it. Kriz will probably conquer it like Grant through Richmond.

$\int\frac{\sqrt{1+x^{2}}}{\sqrt{10-x^{3}}}dx$
I almost got it by letting

$u=\sqrt{1+x^2}$

which results in something nasty that you put a trig sub in for and it looks easy...it is but then WHAM! not anymore...try it for yourself =(

4. Mmm, no, I think there's no antiderivative for this one.

But the user NonCommAlg is better than me. He may give an opinion too.