How do you integrate int x^4ln2x∫x4ln2x by integration by parts method?

1 Answer
Narad T.
Oct 18, 2016

intx^4ln2xdx=(x^5lnx)/5-x^4/4+C∫x4ln2xdx=x5lnx5−x44+C

Explanation:

Let u(x)=ln2xu(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

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