Integration-Explanation Please?

[Justabeginner]

Homework Statement

The problem asks me to integrate this: $∫(1/x) * (e^{-2 log [3, x]})) dx$
I don't know how to write log base 3 x, so that is how I entered it. Sorry if that was confusing. :S

Homework Equations

The Attempt at a Solution

The correct answer is $(-log(3))/(2x^{2/(log 3)}) + C$

But I have no clue how that was even the answer. Guidance is much appreciated. Thank you.

 

[CAF123]

Rewrite $\log_3 x$ in terms of the natural logarithm: $$\log_ax = \frac{\ln x}{\ln a}$$

 

[Justabeginner]

So I'd get int (1/x) * (e^(-2 lnx/ln3)) dx? And is the integral of e to any power that expression times the exponent?

 

[NascentOxygen]

Only if the exponent is a constant (i.e., does not contain any x terms).

You'll simplify things by using the fact that

e[/size]^log_e(x) = x[/size]

 

[CAF123]

So I'd get int (1/x) * (e^(-2 lnx/ln3)) dx? And is the integral of e to any power that expression times the exponent?

No. You have this $$\int \frac{1}{x} \exp \left(-2 \frac{\ln x}{\ln 3}\right)\,dx$$ There is an obvious substitution you can make now.

 

[Justabeginner]

I'm not sure if you can take the integral of 1/x and multiply it by the integral of the ln part?

 

[SammyS]

I'm not sure if you can take the integral of 1/x and multiply it by the integral of the ln part?

No, you can't.

You have to use some additional properties of logarithms and exponential functions to get a somewhat simple integrand.

I say somewhat simple because it will just be x to a power -- but the power is pretty weird looking.

 

[Justabeginner]

Oh! I see. Thank you.

 

[vela]

Since 1/x is the derivative of ln x, you might try the substitution u = ln x as well.