interesting integrals
$$\int \sqrt{\frac{x}{1-x^3}}dx$$
$$\int \frac{1}{\ln x}+\ln(\ln x) dx$$

Question

interesting integrals
$$\int \sqrt{\frac{x}{1-x^3}}dx$$
$$\int \frac{1}{\ln x}+\ln(\ln x) dx$$

xapproachesinfinity · Answer

Second I solved it!
first still beats me

serenity74 · Answer

integrals = cringes

xapproachesinfinity · Answer

any hint for that first one @imqwerty

imqwerty · Answer

m thinking :) 
i havn't learned integrals in depth yet in institute tho

solomonzelman · Answer

$\color{#000000 }{ \displaystyle \int \frac{1}{\ln x}+\ln (\ln x)~dx  }$

z=ln(x)
dz = dx/x    -->    x dz = dx    -->    e^z du =dx

$\color{#000000 }{ \displaystyle \int \left(\frac{e^z}{z}+e^z\ln  z\right)~dz  }$

$\color{#000000 }{ \displaystyle \int \left(\sum_{k=0}^{\infty}\frac{z^{k-1}}{k!}\right)dx+\int\left(e^z\ln  z\right)~dz  }$

the first is series, ez.
second probably integ. by parts

solomonzelman · Answer

nope, same thing,

$\color{#000000 }{ \displaystyle \int\left(e^z\ln  z\right)~dz  =e^z \ln z-\int\frac{e^z}{z}dz}$

solomonzelman · Answer

and that same series.

solomonzelman · Answer

$\color{#000000 }{ \displaystyle \int\sqrt{\frac{x}{1-x^3}}dx}$

$\color{#000000 }{ \displaystyle \int\frac{\sqrt{x}}{\sqrt{1-x^3}}dx}$

$\color{#000000 }{ \displaystyle \int\frac{\sqrt{x^4}}{\sqrt{x^3-x^6}}dx}$

will assume x is real,

$\color{#000000 }{ \displaystyle \int\frac{x^2}{\sqrt{x^3-x^6}}dx}$

$\color{#000000 }{ \displaystyle z=x^3}$
$\color{#000000 }{ \displaystyle (1/3)dz=x^2~dx}$

$\color{#000000 }{ \displaystyle \int\frac{1}{3\sqrt{z-z^2}}dz}$

so far..

xapproachesinfinity · Answer

Solo there a neat solution to second without series

xapproachesinfinity · Answer

first one don't know thus far, i did think of using euclidean division of x/1-x^3

solomonzelman · Answer

trying to think of it...

xapproachesinfinity · Answer

gotta go, good luck in solving them

solomonzelman · Answer

By parts...
$\color{#000000 }{ \displaystyle \int \ln(\ln x)dx=}$

$\color{#000000 }{ \displaystyle x\left(\ln(\ln x)\right)-\int x\left(\frac{1}{x\ln x}\right)~dx}$

So,

$\color{#000000 }{ \displaystyle \int \frac{1}{\ln x}+\ln(\ln x)dx=}$

$\color{#000000 }{ \displaystyle \int \frac{1}{\ln x}dx+\int \ln(\ln x)dx=}$

$\color{#000000 }{ \displaystyle \int \frac{1}{\ln x}dx+x\left(\ln(\ln x)\right)-\int x\left(\frac{1}{x\ln x}\right)~dx=x\ln(\ln x)}$

solomonzelman · Answer

there, @xapproachesinfinity

solomonzelman · Answer

well, +C, but anyway ...

solomonzelman · Answer

As far as the second integral goes

$\color{#000000 }{ \displaystyle \int \frac{1}{\sqrt{z-z^2}}dz  }$

$\color{#000000 }{ \displaystyle \int \frac{1}{\sqrt{-z^2+z-\frac{1}{4}+\frac{1}{4}}}dz  }$

$\color{#000000 }{ \displaystyle \int \frac{1}{\sqrt{-\left(z-\frac{1}{2}\right)^2+\frac{1}{4}}}dz  }$

w=z-1/2
dw=dz

$\color{#000000 }{ \displaystyle \int \frac{1}{\sqrt{-w^2+\frac{1}{4}}}dw  }$

$\color{#000000 }{ \displaystyle \int \frac{1}{\sqrt{\frac{1}{4}-w^2}}dw  }$

$\color{#000000 }{ \displaystyle  w=\frac{1}{2}\sin \theta   }$		$\color{#000000 }{ \displaystyle  dw=\frac{1}{2}\cos \theta   }$

$\color{#000000 }{ \displaystyle \int \frac{\frac{1}{2}\cos \theta}{\sqrt{\frac{1}{4}-\frac{1}{4}\sin^2\theta }}d\theta   }$

$\color{#000000 }{ \displaystyle \int d\theta   }$

xapproachesinfinity · Answer

you go the second excellent!

xapproachesinfinity · Answer

got*

solomonzelman · Answer

Thanks;)

solomonzelman · Answer

didn't put 1/3 there but I hope that is the smallest error I had there.

solomonzelman · Answer

that's referring to the 1st prob.
I like the second problem more tho:) Tnx for posting the problems!

xapproachesinfinity · Answer

np:)

imqwerty · Answer

:) for 1st question-
$\large t=x^{\frac{3}{2}}$  made life easier<33