Ragib

How about [math]\int^1_0 \frac{\ln (1+x)}{x} dx[/math] .

[math]\int \frac{1}{x\sqrt{x}\sqrt{x-1}} dx[/math] Let [math]x=\cosh^2 u[/math], then [math]dx = 2 \cosh u \sinh u du[/math]. [math]\int \frac{2\cosh u \sinh u}{\cosh^3 u \sinh u} du[/math] = 2 \int \sech^2 u du [math]= 2\tanh u + C [/math] [math]= 2\tanh (\cosh^{-1} \sqrt{x} ) + C [/math] EDIT: LaTeX was not working for those two lines, so I ommitted the tags so people could see what I meant to type, and hopefully someone can correct my 'LaTeX syntax error'. (Fixed the LaTeX for you. It turns out there's no \arccosh command in LaTeX, so I had to make do with cosh-1. Oh well. -- Cap'n)

Very simple and fast program that works up to 5011 digits. Extremely useful. Bcalc.zip

But possible with a slightly more powerful geometry of origami

Umm well first, its df/dx, and its as h -> 0. I have no idea what your method is trying to show, but that only works for polynomials anyway. The OP wants a general one for all functions.

Well it says it has to be proven for all positive values, and I can't say the answers have to be the same, so I would assume they can be separate values.

Let [math]x=u^2[/math], and do as Bignose said. The integral becomes [math]2\int \frac{u^3}{1+u} du[/math], which with some easy polynomial division will get you [math]\frac{u^3}{1+u}=u^2-u+1 -\frac{1}{1+u}[/math]. Split the integral in two, you get: [math]2(\int u^2 -u +1 du -\int \frac{1}{1+u} du[/math]. The first one is easy, reverse power rule. The second, do some substitution. So you should get [math]2(\frac{u^3}{3} - \frac{u^2}{2} + u -\log_e |u+1|)[/math]. Substitute [math]u=\sqrt {x}[/math] back in EDIT: Btw I checked the answers, its correct

I dont want to think of it this way, but deep down I think the gloryious days of science are gone, when the recluse in his shed, or the small time amatuer can make a decent contribution. In mathematics its still possible, with some genius, but these days with physics you need at least 3 now.

My bad, I guess I have seen that equation before then Sorry, My Brains Been farting alot recently..

Unless your saying [math]\frac{a}{2} dt[/math] is equal to zero, that equation is incorrect. In other words, velocity is equal to zero. Otherwise Its wrong. I've never seen that equation before.

Well, heres my best shot. Ok, well the rule applies to composite functions, say f(x)/g(x). When we sub in values we get an indeterminate form. We only want the RATIO of the 2 functions. So say 5x/x, x appraoches zero, 0/0. But since we only want the ratio, we could instead get an approximation that, in the limit, is exact. Our approximation is our tangent line The tangent line has the same value as the point it touches. So basically we found the ratio of the values at the tangents, which is what L'hopitals rule wants. I hope I explained that well...

Did you get it?

Well since there is both sin and cos in one equation, all i can think of is t substitution. Let t=tan(x/2), since you know the the expansion for tan(x+y), let x=x/2 and y=x/2, that way tan x= 2t/(1-t^2). Right a right triangle, set 2t the opposite side, 1-t^2 the adjacent, use pythagoras for the remaining sides. Then the opp/hypotenuse ratio is sin x, and do the same for cos. Make the substuitions for cos and sin into the last equation, solve for t and you can get a quadratic equation, which is easy to solve. Good luck

Anyone? Come on, at least the 2nd problem...

I don't think that is correct sinisterwolf, can you give me your values for x y and z seperately? That doesn't look right..