Evaluate the integral ((lnx)^2)/(x^3) from 1 to 5?

Question

I keep trying to find the right answer, but whatever I do doesn't work. Thanks for the help!

kb · Accepted Answer

Using integration by parts...

Let u = (ln x)^2, dv = (1/x^3) dx
du = 2(ln x) dx/x, v = (-1/2) x^(-2).

So, ∫ (ln x)^2 dx/x^3
= (-1/2) (ln x)^2/x^2 - ∫ -(ln x) dx/x^3
= (-1/2) (ln x)^2/x^2 + ∫ (ln x) dx/x^3.

Now, let u = ln x, dv = (1/x^3) dx
du = dx/x, v = (-1/2) x^(-2).

So, ∫ (ln x)^2 dx/x^3
= (-1/2) (ln x)^2/x^2 + [(-1/2) (ln x)/x^2 - ∫ (-1/2) dx/x^3]
= (-1/2) (ln x)^2/x^2 + (-1/2) (ln x)/x^2 - (1/4)/x^2 + C
= (-1/4) x^(-2) [2(ln x)^2 + 2 (ln x) + 1] + C.
--------------
Hence, ∫(x = 1 to 5) (ln x)^2 dx/x^3 
= (-1/4) x^(-2) [2 (ln x)^2 + 2 (ln x) + 1] {for x = 1 to 5}
= (-1/4) 5^(-2) [2 (ln 5)^2 + 2 (ln 5) + 1] + 1/4
= (1/50) [12 - (ln 5)^2 -  ln 5].

(Double checked on Wolfram Alpha.)

I hope this helps!

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Evaluate the integral ((lnx)^2)/(x^3) from 1 to 5?

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