Question

I really need help with this plz i) Find `d/dx(x^3lnx)` ii) Hence find `intx^2lnxdx` ?

i,ve done the first part but i cant do the second
very badly stuck

Answer 1

I get `(x^3(3lnx-1))/9+C`.

Explanation:

Given: `intx^2lnx\ dx`

Apply integration by parts, `intu \ dv=uv-intv \ du`.

Here, I'll let `u=lnx,dv=x^2`

`:.du=1/x \ dx,v=1/3x^3` (no need `C` yet)

So, going back to our original integral, we get

`int(x^2lnx) \ dx=lnx*1/3x^3-int(1/3x^3*1/x) \ dx`

`=(x^3lnx)/3-int(x^2/3) \ dx`

`=(x^3lnx)/3-1/3intx^2 \ dx`

`=(x^3lnx)/3-1/3*x^3/3+C`

`=(x^3lnx)/3-x^3/9+C`

`=(3x^3lnx-x^3)/9+C`

`=(x^3(3lnx-1))/9+C`

Answer 2

`d/dx (x^3lnx) = x^2+3x^2lnx`

`int \ x^2lnx \ dx = 1/3x^3lnx - x^3/9 + C`

Explanation:

Part (i)

We seek:

`d/dx (x^3lnx)`

We can use the product rule, to get:

`d/dx (x^3lnx) = (x^3)(d/dxlnx))+(d/dxx^3)(lnx)`
`" "= (x^3)(1/x)+(3x^2)(lnx)`
`" "= x^2+3x^2lnx`

Part (ii)

"Hence", indicates we should probably use the earlier result:

`d/dx (x^3lnx) = x^2+3x^2lnx`

Integrate (term by term) wrt `x` (omitting the constant of integration), we get

`x^3lnx = int \ x^2 \ dx + int \ 3x^2lnx \ dx`

The first integral is trivial (power rule), the second integral is the result we seek, thus:

`x^3lnx = x^3/3 + 3 \ int \ x^2lnx \ dx`

And we can rearrange:

`3 \ int \ x^2lnx \ dx = x^3lnx - x^3/3`

And incorporating the constant of integration:

`int \ x^2lnx \ dx = 1/3x^3lnx - x^3/9 + C`