Question

How do you integrate `int x^3lnx` by integration by parts method?

Answer 1

`intx^3ln(x)dx=(x^4(4ln(x)-1))/16+C`

Explanation:

Integration by parts takes the form:

`intudv=uv-intvdu`

So, for the integral `intx^3ln(x)dx`, we see this is `intudv` and let:

`{(u=ln(x)" "=>" "du=1/xdx),(dv=x^3dx" "=>" "v=x^4/4):}`

Thus:

`intx^3ln(x)dx=uv-intvdu=(x^4ln(x))/4-intx^4/4 1/xdx`

Simplifying the integral:

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

`intx^3ln(x)dx=(x^4ln(x))/4-1/4x^4/4+C`

`intx^3ln(x)dx=(x^4ln(x))/4-x^4/16+C`

`intx^3ln(x)dx=(x^4(4ln(x)-1))/16+C`

Answer 2

`int x^3lnxdx=1/4x^4lnx - x^4/16 + C`

Explanation:

The formula for integration by parts is:
`intu(dv)/dxdx = uv - intv(du)/dxdx`

Essentially we would like to find one function that simplifies when differentiated, and one that simplifies when integrated (or is at least integrable).

Let `{(u=lnx, => ,(du)/dx=1/x),((dv)/dx=x^3,=>,v =x^4/4 ):}`

So IBP gives:

`int (lnx)(x^3)dx=(lnx)(x^4/4) - int (x^4/4)(1/x)dx`
`:. int x^3lnxdx=1/4x^4lnx - 1/4int x^3dx`
`:. int x^3lnxdx=1/4x^4lnx - 1/4(x^4/4) + C`
`:. int x^3lnxdx=1/4x^4lnx - x^4/16 + C`