i need help u-substituting xlnxdx?

Question

how do you do the u-substitution for this..i mean like what is u? du? dv? v? how could u figure out which is which?
i thought
u= x dv= lnxdx
du= xdx v= 1/x

Answer 1

To integrate x ln x, you need to understand the process for integrating ln x - it's basically the same thing. It's clear that you don't understand integration by parts yet, since your u and dv do not match your du and v respectively.

Note that du is (du/dx) dx and v = ∫dv (but without the constant).
So if u = x and dv = ln x dx, then du = dx and v = ∫ln x dx = x ln x - x (we'll show this in a minute).

So, consider ∫ln x dx. We integrate by parts, choosing:
u = ln x, dv = 1 dx
and so du = (d/dx ln x) dx = (1/x) dx, v = ∫1 dx = x.
So we get
∫ln x dx = ∫u dv = uv - ∫v du
= (ln x) (x) - ∫x (1/x) dx
= x ln x - ∫1 dx
= x ln x - x + c.

Now we have this, we use the same trick to integrate x ln x:
∫x ln x dx: let u = ln x, dv = x dx
Then du = 1/x dx, v = x^2/2
so we have
∫x ln x dx = (x^2/2) ln x - ∫(x^2/2)(1/x) dx
= (x^2 ln x) / 2 - ∫(x/2) dx
= (x^2 ln x) / 2 - x^2 / 4 + c.

Answer 2

What you initially wrote in the question part is almost right...
u = lnx ... dv = x dx
du = dx/x .... v = 1/2*x^2

uv - int v*du
1/2x^2*lnx - 1/2 int (x^2/x *dx)
1/2x^2*lnx - 1/2 int (x dx)
1/2*x^2*lnx - 1/4*x^2 +c

you decided to chose u as lnx and dv as x dx, because u is most likely a power of x and dv is usually the other thing...

Answer 3

let
u = lnx
du = dx/x

and
v = x dx
dv = x²/2

so:

int(udv) = uv - int(vdu)
=x ln(x) - int(x²/2 * dx/x)
= x ln(x) - (1/2) int(x dx)
= x ln(x) - (1/4) x² + C

In general if you have x * f(x), were f(x) could be sin(x), exp(x), you would pick u to be x, when f(x) is ln(x), you pick u=ln(x) becuase the derivative of ln(x) is just 1/x.

Also, in your work you have dv = lnx dx v = 1/x

recal that int(1/x) = ln x and int(ln x) = xlnx - x
you can prove that if you let u = ln x and dv = dx, another speical case.

you pick dv = dx if you are trying to integrate ln(x), arctan(x), arcsin(x) ...

Answer 4

its just useful to memorize the derivative of ln x it comes up too often
use parts
u = lnx dv=x ( this way is better because you dont know the integral of ln x )
du=1/x v=x^2/2
then use the formula
u*v - S v du

the S is integral
good luck

Answer 5

du = dx. So,
uv- int(1/x dx)