Tricky integral:
$$\int e^{-2x}\left[(e^x)^2 +(x^x)^2\right]\ln x dx$$
No Googling allowed :)

Question

Tricky integral:
$$\int e^{-2x}\left[(e^x)^2 +(x^x)^2\right]\ln x dx$$
No Googling allowed :)

solomonzelman · Answer

u=e^x

anonymous · Answer

haha no googling allowed,what about wofram????

solomonzelman · Answer

I mean u=e^2x

anonymous · Answer

Go right ahead @Jonask, just stay away from Math Stack Exchange!

solomonzelman · Answer

$$\int\limits_{ }^{ }e^{-2x}2e^{2x}~dx$$$$\int\limits_{ }^{ }2~dx$$and it is 2x?

solomonzelman · Answer

Ohh the ln. sorry

solomonzelman · Answer

it would be integral of 2ln(x) I guess. After simplifying the Es

anonymous · Answer

$$\int  \ln x dx+\int \frac{x^{2x}}{e^{2x}}\ln xdx$$

anonymous · Answer

@SolomonZelman your simplification isn't right. The first term should simplify to 1 (multiplied by $\ln x$).

solomonzelman · Answer

but in your brackets isn't it just 2e^2x ?

hartnn · Answer

i see the derivative of x^x
in the function

solomonzelman · Answer

and e^2x * e^-2x= e^(-2x + 2x) = e^0 = 1
So,
it would be with a 2.

hartnn · Answer

which is x^x (1+ln x)

anonymous · Answer

Second term is $\large x^x$, not $\large e^x$...

Yes, @hartnn is on the right track. It's a matter of coming up with the right sub now.

solomonzelman · Answer

ohh.. I am blind again

ganeshie8 · Answer

$$\large   \int e^{-2x}\left[(e^x)^2 +(x^x)^2\right]\ln x dx = \int (1 + e^{2x(\ln x - 1)}) \ln x  $$ 

sub  2x(lnx-1) = u ?

anonymous · Answer

Not what I would have suggested @ganeshie8, but it might still lead to a solution.

hartnn · Answer

first integral is ln x
u = e^(-2x) x^(2x) for the 2nd integral

hartnn · Answer

du = 2 e^(-2x) x^(2x) ln x

hartnn · Answer

*dx

hartnn · Answer

$\int  \ln x dx+\int \frac{x^{2x}}{e^{2x}}\ln xdx \u = \frac{x^{2x}}{e^{2x}} \ du =2\frac{x^{2x}}{e^{2x}} dx \\int  \ln x dx+\int \frac{du}{2}$

and the rest is history :P

ganeshie8 · Answer

Clever! :)

anonymous · Answer

Nice work! Honorable mention @Jonask, you had the right approach as well

hartnn · Answer

was just trying
u =x^(2x) before!
then realized how simple u = e^(-2x)x^(2x) made this problem :D

hartnn · Answer

thanks for the good question! :)

anonymous · Answer

i deleted i thought it was not gonna work

anonymous · Answer

the rest is history,nice! @hartnn

hartnn · Answer

lol, too lazy to write out those easy steps :P

hartnn · Answer

bdw, integral of ln x would be solved using integration by parts

hartnn · Answer

***
$\large du =2\frac{x^{2x}}{e^{2x}} \ln xdx$