K!!

[math]{{x}^{\sin x}}=\exp \left( \sin (x)\ln x \right)=\exp \left( \frac{\sin x}{x}\cdot x\ln x \right)\to 1[/math] as [math]x\to0,[/math] since [math]\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}=1[/math] and [math]\underset{x\to 0}{\mathop{\lim }}\,x\ln x=0.[/math]

Simplify applyin' properties: [math](p\implies q)\implies[p\implies(\sim q\wedge p)].[/math]

Why do you need an integrating factor? Your equation is separable.

Here's another one: http://www.fmat.cl (Spanish.)

Cap'n Refsmmat... so... aren't you gonna set the math button?

Okay well, after series expansion we have: [math]\begin{aligned} \int_0^1 {\frac{{\ln (1 + x)}} {x}\,dx} &= \sum\limits_{k = 0}^\infty {\frac{{( - 1)^k }} {{k + 1}}\left\{ {\int_0^1 {x^k \,dx} } \right\}}\\ &= \sum\limits_{k = 0}^\infty {\frac{{( - 1)^k }} {{(k + 1)^2 }}}\\ &= \frac{{\pi ^2 }} {{12}}. \end{aligned}[/math]

It is a well known one; its solution is based on series. The answer is [math]\frac{{\pi ^2 }} {{12}}[/math]. Can you edit your post with another integral?

Okay well [math]\int {\frac{1} {{x\sqrt {x^2 - x} }}\,dx} = \frac{{2\sqrt {x^2 - x} }} {x} + k[/math]. Here's my approach: Set [math]x=\frac1u,[/math] the integral becomes [math]- \int {\frac{1} {{\sqrt {1 - u} }}\,du} = 2\sqrt {1 - u} + k[/math]. The rest follows. -- Post your integral now.

Do you mean [math]x\,\frac{{dy}} {{dx}} = y + xy^2[/math]? This is not a linear ODE, so you can't solve it with integratin' factor method. This is actually a Bernoulli's Differential Equation.

Well, I'm not actually requesting that. I use MathType to type LaTeX codes, it's faster.

I'd like to initiate a kind of Integral Marathon. This is simple, person who solves a problem must receive a confirmation whether the answer is correct or not. In case where the answer is correct, solver may post the next integral. (Of course, indefinite & finite integrals are allowed.) Let's start with an easy one: Solve [math]\int {\frac{1} {{x\sqrt {x^2 - x} }}\,dx}[/math].

Yes, this is a good point. Is there a LaTeX manual designed? I can design one for new people.

Thanks for considering my suggestion Cap'n Refsmmat

It is for save time. It's quickly when you're helpin' someone to solve a problem. (I'm not askin' to set all symbols, it's enough by settin' [math][/math] tags.)

The faster way to solve this, it's by settin' [math]u^2=4x^2-9[/math]. The rest follows.

Sorry, I don't get very well what you mean.

Have you people considered to design a LaTeX button which generates [math][/math] tags instantly without writtin'em?

Actually, this is the nice method to tackle such integrals. It takes less time.

[math]\int {\frac{{dx}} {{\sqrt {1 - x^2 } \cdot \arcsin x}}} = \int {\frac{{(\arcsin x)'}} {{\arcsin x}}\,dx} = \ln \left| {\arcsin x} \right| + k[/math].

I've been teachin' me about one year and it's pretty cool. You have to be patient to understand complex things.