Pls solve ! (10 points )?

Let w = ln x, dw = dx/x, the integrand becomes:

w^2 e^(2w) dw.

Now use integration by parts:

u = w^2, dv = e^(2w) dw, du = 2w dw, v = (1/2)e^(2w).

You have uv - integral of v du

= (1/2)w^2 e^(2w) - integral of [w e^(2w) dw]

Use integration by parts again: u2 = w, dv2 = e^(2w) dw, du2 = dw, v2 = (1/2)e^(2w). Now you have:

(1/2)w^2 e^(2w) - (1/2)w e^(2w) + integral of (1/2)e^(2w) dw

= (1/2)w^2 e^(2w) - (1/2)w e^(2w) + (1/4)e^(2w)

= (1/2)x^2 [ln(x)]^2 - (1/2)x^2 ln(x) + (1/4)x^2.

Now plug in x = e and 0, obtaining:

(1/2)e^2 - (1/2)e^2 + (1/4)e^2 - 0 = (1/4)e^2.

Since my answer (even before the substitution of x values) is quite different from cidyah's answer, one of us has made an algebra mistake. Good exercise for you to find out which answer is correct.

∫ x^3 (ln x)^2 dx

Integrate by parts

dv= x^3 dx ; v= (1/4) x^4

u = (ln x)^2 ; du = 2 (ln x / x) dx

∫ u dv = u v - ∫ v du

∫ x^3 (ln x)^2 dx = (1/4) (ln x)^2 x^4 - (1/4) ∫ x^4 (2 ln x / x) dx

∫ x^3 (ln x)^2 dx = (1/4) (ln x)^2 x^4 - (1/2) ∫ x^3 ln x dx -------------- (1)

Integrate ∫ x^3 ln x dx by parts

dv=x^3 dx; v= (1/4) x^4

u = ln x ; du = 1/x dx

∫ u dv = u v - ∫ v du

∫ x^3 ln x dx = (1/4) ln x x^4 - (1/4) ∫ x^4 (1/x) dx

∫ x^3 ln x dx = (1/4) ln x x^4 - (1/4) ∫ x^3 dx

∫ x^3 ln x dx = (1/4) ln x x^4 - (1/4)(1/4) x^4

∫ x^3 ln x dx = (1/4) ln x x^4 - (1/16) x^4

substitute this into (1)

∫ x^3 (ln x)^2 dx = (1/4) (ln x)^2 x^4 - (1/2) ( (1/4) ln x x^4 - (1/16) x^4 )

∫ x^3 (ln x)^2 dx = (1/4) (ln x)^2 x^4 - (1/8) ln x x^4 + (1/32) x^4

Let F(x) = (1/4) (ln x)^2 x^4 - (1/8) ln x x^4 + (1/32) x^4

substitute the upper limit e

F(e) = 1/4) e^4 - (1/8) e^4 + (1/32) e^4 =(5/32) e^4

substitute the lower limit 1

F(1) = 1/32

subtract:

F(e)-F(1) = (5/32) e^4 - 1/32

Put lnx = t

So dx = e^t dt

new Limit (0-->1)

So integeration become

I = §(e^4t)(t^2)dt

Use ilate with u=t^2 and v=e^4t

And put limits

You will get

I= (5*e^4 - 1)/32