Stuck on a question using integration of parts.
integral sign x^2ln4x dx
u=ln4x v=1/3x^3
du= dv=x^2
S u dv = uv - S v du
ln4x( )-S1/2x^3( )
I don't know how to find du above? Is it 1/x(4x)??
Any help would be great, thanks.
hi bradycat
you can look at it like this:
if u=ln(4x) then the derivative of u with respect to x is
du/dx=4/4x=1/x
multiply both sides by dx and you get:
du=dx/x
also dv should be:
v=1/3 x^3
dv/dx=x^2
dv=x^2 dx.
don't forget your dx.
Slight misunderstanding, the OP had already integrated. (But = c wants to be + c of course.)
As confirmed by Hugal. But here's a pic, just in case it helps...
... where (key in spoiler) ...
Spoiler:
__________________________________________________ __________
Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
Balloon Calculus Drawing with LaTeX and Asymptote!