Evaluate the integral? : # int_1^e x^5 (lnx)^2 dx #

Question

Evaluate the integral? : # int_1^e  x^5  (lnx)^2  dx #

Elijah Abram · Accepted Answer

# int_1^e \ x^5 \ (lnx)^2 \ dx = (13e^6 -1)/108 #

We want to find: # I = int_1^e \ x^5 \ (lnx)^2 \ dx # We can use integration by Parts Let # { (u,=(lnx)^2, => (du)/dx,=(2lnx)/x), ((dv)/dx,=x^5, => v,=x^6/6 ) :}# Then plugging into the IBP formula: # int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx # gives us # int_1^e \ (lnx)^2(x^5) \ dx = [(lnx)^2(x^6/6)]_1^e - int_1^e \ (x^6/6)((2lnx)/x) \ dx # # :. I = 1/6(e^6ln^2e - ln^21) - 1/3 \ int_1^e \ x^5lnx \ dx # # " "= e^6/6 - 1/3 \ int_1^e \ x^5lnx \ dx # So now let us look at this next integral: # J = int_1^e \ x^5lnx \ dx # Again we can use integration by Parts Let # { (u,=lnx, => (du)/dx,=1/x), ((dv)/dx,=x^5, => v,=x^6/6 ) :}# and plugging into the IBP formula gives us # int_1^e \ (lnx)(x^5) \ dx = [(lnx)(x^6/6)]_1^e - int_1^e \ (x^6/6)(1/x) \ dx # # :. J = 1/6(e^6lne-ln1) - 1/6 \ int_1^e \ x^5 \ dx # # " " = e^6/6 - 1/6 \ [x^6/6]_1^e # # " " = e^6/6 - 1/36 \ (e^6-1) # Combining these results we get: # I = e^6/6 - 1/3 (e^6/6 - 1/36 \ (e^6-1)) # # \ \ = e^6/6 - 1/3 (e^6/6 - e^6/36+1/36) # # \ \ = e^6/6 - 1/3 (5/36e^6 +1/36) # # \ \ = e^6/6 -5/108 e^6 -1/108 # # \ \ = 13/108 e^6 -1/108 # # \ \ = (13e^6 -1)/108 #

Emilia Aguilar · Answer

undefined To evaluate the integral ∫(1 to e) x^5(lnx)^2 dx, you can use integration by parts.

Let u = (lnx)^2 and dv = x^5 dx.
Then, differentiate u to get du and integrate dv to get v.

u = (lnx)^2
dv = x^5 dx

Differentiating u:
du = 2(lnx)(1/x) dx
du = (2/x)lnx dx

Integrating dv:
v = (1/6)x^6

Now, apply the integration by parts formula:

∫ u dv = uv - ∫ v du

∫(1 to e) x^5(lnx)^2 dx = [(1/6)x^6(lnx)^2] from 1 to e - ∫(1 to e) (1/6)x^6 (2/x)lnx dx

Evaluate the integral from 1 to e:

= [(1/6)e^6(ln(e))^2 - (1/6)(1)^6(ln(1))^2] - ∫(1 to e) (1/6)x^5(2lnx) dx

= [(1/6)e^6(1)^2 - (1/6)(0)^2] - ∫(1 to e) (1/3)x^5(lnx) dx

= (1/6)e^6 - 0 - (1/3)∫(1 to e) x^5(lnx) dx

Now, you can integrate ∫ x^5(lnx) dx using integration by parts again or by recognizing it as a tabulated integral. The final result will be:

= (1/6)e^6 - (1/3)[(1/6)e^6 - (1/3)∫(1 to e) x^4 dx]

= (1/6)e^6 - (1/3)[(1/6)e^6 - (1/3)(1/5)e^5 - (1/3)(1/5)]

= (1/6)e^6 - (1/3)[(1/6)e^6 - (1/15)e^5 - (1/15)]

= (1/6)e^6 - (1/18)e^6 + (1/45)e^5 + (1/45)

= (4/9)e^6 + (1/45)e^5 + (1/45)