Is the Antiderivative of 2(1 + ln(x))(x^x)^2 Solvable?

[roam]

[SOLVED] Unsolvable Problem?

Hello!

How can we find the antiderivative of 2(1 + lnx)(x^x)^2


As it stands it is not algebraically solvable.
This question is from my calculus' book, but the solutions are not provided...

Does anyone know of any mthod we can use to integrate this?

 

[uman]

Substitution. u=(x^x)^2.

 

[HallsofIvy]

Oh, that is clever!

 

[ice109]

actually that is the antiderivative x^(2x)

 

[uman]

Yep.

\int 2(1+\log x)(x^x)^2\, dx: u=(x^x)^2; by logarithmic differentiation we find du=2(x^x)^2(\log x + 1) so the integral is \int\, du = u = (x^x)^2 = x^{2x}.

 

[ice109]

so how did you figure out that substitution?

 

[LukeD]

If I had to guess, I'd have to say it was either
1. a barter for his soul with some demon/deity
2. divine intervention
3. ? can't think of any other possibility

Seriously, that was clever

 

[HallsofIvy]

I suspect that he realized that, since the derivative of x^x involves a logarithm, and there was a logarithm in the problem, he should try something like u= x^x, tried it, saw that it didn't quite work and then played with it until he found something that did.

Two many math students seem to think that the way you solve a math problem is to stare at the paper until the solution springs, like Venus, full grown from your forehead.

 

[ice109]

I suspect that he realized that, since the derivative of x^x involves a logarithm, and there was a logarithm in the problem, he should try something like u= x^x, tried it, saw that it didn't quite work and then played with it until he found something that did.

Two many math students seem to think that the way you solve a math problem is to stare at the paper until the solution springs, like Venus, full grown from your forehead.

maybe but writing simply for the sake of writing isn't effective either.

 

[uman]

I dunno, I just saw it and tried it and voilà, it worked.

I knew the answer was (x^x)^2 before I solved the problem, which may or may not have helped, but probably did.

 

[uman]

Also I left out a dx in my first post.

 

[uman]

Also (and sorry for the triple post!) u=x^x would have worked just as well.

 

[roam]

O.K...

Here is my working, by using substitution:


Let I = ∫2(1+lnx)(x^x)2 dx

Let u = x^x

∴ lnu = lnx^x

∴ lnu = xlnx

∴ 1/u × dudx = x×1/x + 1×lnx

∴ du/dx = u ×(1+lnx)

∴ du = x^x^×(1+lnx)dx

∴ dx = du / x^x^×(1+lnx)

∴ I = ∫2(1+lnx)(x^x)^2 dx

= ∫2(1+lnx)(x^x)^2{du / [x^x×(1+lnx)]}

= ∫2 x^xdu

= ∫ 2u du

= u² + c

= (x^x)^2 + c

= x^2x + c


I didn't know if it was correct but now since you have the same solution as me then it must be right...

Many thanks.

 

[uman]

Yep, that seems right although it was hard to follow somewhat because you didn't use LaTeX...

 

[cristo]

That makes for quite an amusing thread title!

 

[Defennder]

I concur.

 

[uman]

Oh man I was just thinking that! You have mah brain, cristo.

 

[chi2cali08]

my calculus teacher taught us this a couple weeks ago. its pretty useful

 

[roam]

Sorry...

I know...about the title...

But it looked unsolvable to me because the original question looked like this: 2(1 + lnx)x^{x^2}

I reckon they forgot to include the brackets. But my teacher then told me to include the brackets and then solve it that way.
2(1 + lnx)(x^x)^2

Thanks.