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.
$\displaystyle u = \ln(4x) \Leftrightarrow u' = \dfrac{1}{x}$
$\displaystyle x^2 \ln(4x) - \int \left(\dfrac{1}{2}x^3 \times \dfrac{1}{x}\right)\ dx$
$\displaystyle = x^2 \ln(4x) - \dfrac{1}{2} \int x^2 \ dx$
You choose to differentiate $\displaystyle x\mapsto \ln 4x$ and integrate $\displaystyle x\mapsto x^2$.
$\displaystyle \int x^2 \ln 4x \textrm{d}x = \frac{x^3}{3}\ln 4x - \int \frac{x^3}{3}\cdot \frac{1}{x} \textrm{d}x = \frac{x^3}{3}\ln 4x -\frac{1}{3} \int x^2 \textrm{d}x$
So :
$\displaystyle \int x^2 \ln 4x \textrm{d}x = \frac{x^3^}{3}\big(\ln 4x - \frac{1}{3}\big) + \lambda$
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!