Question

How do you integrate `int x^4ln2x` by integration by parts method?

Answer 1

`intx^4ln2xdx=(x^5lnx)/5-x^4/4+C`

Explanation:

Let `u(x)=ln2x` so `u'(x)=2/(2x)=1/x`
`v'(x)=x^4` so `v(x)=x^5/5`
Integration by parts is
`intu(x)v'(x)dx=u(x)v(x)-intu'(x)v(x)dx`
Applying this we get
`intx^4ln2xdx=(x^5lnx)/5-intx^4dx/x`
`=(x^5lnx)/5-intx^3dx`
`=(x^5lnx)/5-x^4/4+C`