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].

[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)

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.

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