int [x^(2x) * (1 + ln(x))] dx

Y

yossarian

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Oct 20, 2006
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Ok this is an integral which I've been stuck on... I think I'm missing a fundamental piece of the theory behind solving this. Any help/explanation would be greatly appreciated.

Integrate:

x^(2x) * (1 + ln(x)) * dx

What is confusing me is that it is not a straight up exponential/power function... it's a combination of both, and I have no clue how to do this. Thanks in advance.

Well according to the integrator, the answer is x^(2*x) / 2

Can someone walk me through the steps please?

Basically, how do you differentiate x^(2x), i guess thats what im asking, and can you even solve this integral with u substitution?

I mean the answer is fine and dandy, but I have no clue what steps would get me there.
 
Re: stuck on another integral

yossarian said:
it's a combination of both
Click to expand...
To me, this is a giant red flag directing you toward your new friend, "Integration by Parts". Have you two met?
 
Actually our teacher sort of glossed over that, I'll see what I can find though...
 
Ok I read up on integration by parts, and I'm still stuck at how to differentiate/integrate x^(2*x)

I haven't dealt with this combo between power/exponential function before... please someone guide me! This is driving me crazy..
 
Re: stuck on another integral

Hello, yossarian!


\(\displaystyle \L\int\)\(\displaystyle x^{^{2x}}(1\,+\,\ln x )\,dx\)

According to the integrator, the answer: \(\displaystyle \,\frac{1}{2}x^{2x}\)
Click to expand...

It turns out to be a "simple" substitution, but one like we've never seen before.


Let \(\displaystyle u \,=\,x^{^{2x}}\) . . . Now find \(\displaystyle dx\) and substitute, right?

To find \(\displaystyle dx\), we use Logarithmic Differentiation.

Take logs: \(\displaystyle \:\ln u \:=\:\ln\left(x^{^{2x}}\right) \:=\;2x\cdot\ln x\)

Then: \(\displaystyle \:\frac{1}{u}\cdot du\:=\:\left[2x\cdot\frac{1}{x}\,+\,2\ln x\right]dx\:=\:2(1\,+\,\ln x)dx\;\;\Rightarrow\;\;dx\:=\:\frac{du}{2(1\,+\,\ln x)u}\)


Substitute: \(\displaystyle \L\:\int\)\(\displaystyle u(1\,+\,\ln x)\,\cdot\,\frac{du}{2(1\,+\,\ln x)u}\)

Simplify: \(\displaystyle \:\frac{1}{2}\L\int\)\(\displaystyle du\)

Integrate: \(\displaystyle \L\:\frac{1}{2}u\,+\,C\)

Back-substitute: \(\displaystyle \L\:\frac{1}{2}x^{^{2x}}\,+\,C\;\;\) . . . ta-DAA!

 
Yep. "Parts" is a bit circular on this one. Bone up on it anyway. :)
 
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