Question

How do you integrate `(x^3)ln(x)dx`?

Answer 1

`x^4/16(4lnx-1)+C`,

OR,

`ln{K(x^4/e)^(x^4/16)}`

Explanation:

We use the Rule of Integration by Parts, which states,

`intuvdx=uintvdx-int[(du)/dxintvdx]dx`

We take `u=lnx, v=x^3rArr(du)/dx=1/x,&, intvdx=x^4/4`

Hence, `I=intx^3lnxdx=x^4/4*lnx-int(1/x*x^4/4)dx`

`=(x^4lnx)/4-1/4intx^3dx=(x^4lnx)/4-x^4/16`

`:. I=x^4/16(4lnx-1)+C`

`I` can further be shortened as under :-

`I=x^4/16(4lnx-1)=x^4/16(lnx^4-lne)=x^4/16ln(x^4/e)`

`:. I=ln(x^4/e)^(x^4/16)+lnK=ln{K(x^4/e)^(x^4/16)}`

Enjoy Maths.!

Answer 2

`1/4 x^4 log_e x-x^4/16`

Explanation:

`d/(dx)(x^{n+1}log_e x)=(n+1)x^n log_e x+x^n`

so

`int x^n log_e x dx = 1/(n+1)x^{n+1}log_e x-int x^n/(n+1)dx+C`

Finally

`int x^n log_e x dx = 1/(n+1)x^{n+1}log_e x-x^{n+1}/(n+1)^2+C`

for `n = 3` we have the result

`1/4 x^4 log_e x-x^4/16 +C`