bleh.. lil integration help T_T (1 Viewer)

[Rush152]

how would i do this question

Integral (x^3 log x) dx


and would someone please tell me what d/dx and dy/dx mean?? lol
(i've never understood it since yr11 and always wrote dashes XD (eg x' = bleh))

 

[Rorix]

d/dx is usually the derivative of a function like d (x^2 + 2/x)/dx sort of thing
dy/dx is y'


and use parts
u=lnx
u'= 1/x
dv= etc.

 

[Shuter]

lol, 2003 SC paper question 1 b hey?

I just did that about an hour ago.

Yes it's just integration by parts, too lazy to post the working, but I got

[(x^4)(4((lnx)-1))]/16

 

[Rush152]

ahhahha yeaah XD
i jus found out i botched the working out big time ><


yaaaay d/dx!! thaanks a lot!! :uhhuh:

 

[Rush152]

another intigration prob ><
(one of my bad topics bleh, gotta go over 3u work..)


in the Patel book there's an example saying
8. int (tan^3 x) dx

and it goes on one of the later steps

= int (tan x . sec^2 x) dx - int (tan x) dx
= int u du ..... where u = tan x

how on earth does that work? i dun geddit T_T

 

[Rorix]

its mathemagics

tan squared = sec squared - 1

 

[tempco]

tan^3x = tan^2x * tanx
tan^2x = sec^2x - 1
tan^3x = (sec^2x - 1)tanx

= sec^2x*tanx - tanx