Integral of 1/(sqrt(x^2 *ln(x)) from e to e^2 No idea.

[Lo.Lee.Ta.]

Integral of 1/(sqrt(x^2 *ln(x)) from e to e^2... No idea. :(

1. ∫e to e^2 of (1/(√x²ln(x)))dx


2. So confused! :(

Okay... So I tried something!

∫e to e^2 of [-2/3(x²*ln(x))^-3/2 * 1/(x + 2xln(x))] |e to e^2


I got the 1/(x+2xln(x)) because I was trying to think of some way to cancel out the
x+2xln(x) that results from the chain rule...

Don't think this is right... If not, what am I supposed to do here?

 

[Ray Vickson]

1. ∫e to e^2 of (1/(√x²ln(x)))dx


2. So confused! :(

Okay... So I tried something!

∫e to e^2 of [-2/3(x²*ln(x))^-3/2 * 1/(x + 2xln(x))] |e to e^2


I got the 1/(x+2xln(x)) because I was trying to think of some way to cancel out the
x+2xln(x) that results from the chain rule...

Don't think this is right... If not, what am I supposed to do here?

For x > 0 we have $\sqrt{x^2 \ln(x)} = x \sqrt{\ln(x)},$ because $\sqrt{x^2} = x$ for x > 0. Change variables to ln(x) = y.

 

[Lo.Lee.Ta.]

Oh. I think I know what you mean. It's not as hard as I thought it would be!

∫e to e^2 of [1/(xsqrt(ln(x)))]

u = ln(x)
du= (1/x)dx

∫e to e^2 [1/(sqrt(u))]du

= ∫ u^(-1/2) du = 2(u)^1/2 |e to e^2

= 2(ln(e^2))^1/2 - 2(ln(e))^1/2

= 2(2)^1/2 - 2(1)^1/2

= 2sqrt(2) - 2 <----Should be answer!

Thanks, Ray Vickson! :D