a quick integration question (1 Viewer)

[Gibbo69er]

hey guys, can anybody go through, with working out how to integrate:

Ln x

 

[FinalFantasy]

hey guys, can anybody go through, with working out how to integrate:

Ln x

int. ln x dx

Consider: y=xlnx-x
dy\dx=lnx+(1\x)x-1
=lnx+1-1=lnx
.: integrate ln x dx=xlnx-x


edit: missed an x as trev mentioned

 

[Trev]

int. ln(x) is done by 'parts', it is an extension 2 method:
int. ln(x) = int. 1.ln(x)= uv - int. (u.dv/dx)dx
Let 1 = du/dx and ln(x) = v.
∴ int. ln(x) = x.ln(x) - int. (x.{1/x})dx
= x.ln(x) - int. (1)dx
= xln(x) - x.

 

[FinalFantasy]

by the way ur not supposed to do integral of lnx in 2unit maths
even if u do u would be given an equation such as the one i wrote there den given steps..

 

[Trev]

int. ln x dx

Consider: y=xlnx-x
dy\dx=lnx+(1\x)x-1
=lnx+1-1=lnx
.: integrate ln x dx=lnx-x

Final Fantasy, you missed out on an 'x' there; it's 'xlnx-x'.

 

[FinalFantasy]

ooops LoL

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[Slidey]

Use the rectangle method if you've been given terminals, Gibbo. Otherwise it's beyond the scope of 2u.

Here is an explanation of the method these monkeys used, though:

Here is the product rule which you should know from 2unit:
d(uv)/dx=u*d(v)/dx + v*d(u)/dx, where u and v are both functions of x.
Let us represent this as: (uv)'=uv'+u'v
Integrate both sides: uv=Int(uv')+Int(u'v)
Move one integral to one side: Int(uv')=uv-Int(u'v)

Now, in the case of lnx, we can represent this as (x)'*lnx
Thus Int(lnx) = Int(lnx * (x')) = x*lnx - Int(1/x * x) = xlnx-x
Because for lnx=(x)'lnx, v'=1, v=x, u=lnx

 

[A2RAYA]

u can use the reduction method if you're given a reduction formula....or you can derive one...but then again you wouldn't need to in 2U