# Quick integration?Watch

Announcements
This discussion is closed.
#1
Is it possible to use integration by subsitution for

∫ x²/(4-x) dx
0
14 years ago
#2
(Original post by endeavour)
Is it possible to use integration by subsitution for

∫ x²/(4-x) dx
Yes. Do you want me to tell you what substitution to make?
0
#3
Argh got it now thanks.
It was just me being stupid and failing to realise (a+b)/c = a/c + b/c
0
14 years ago
#4
(Original post by endeavour)
Find ∫ x²/(4-x) dx
Let u = 4 - x --> du/dx = -1 --> dx/du = -1 --> dx = -1 du

Hence: ∫ x²/(4-x) dx
= ∫ [(4 - u)(4 - u)]/u . (-1) du
= - ∫ (u^2 - 8u + 16)/u du
= - ∫ u - 8 + 16(1/u) du
= - [(u^2)/2 - 8u + 16lnu] + k
= 8u - (u^2)/2 - 16lnu + k
= 8(4 - x) - [(4 - x)^2]/2 - 16ln(4 - x) + k
= 32 - 8x - [(4 - x)^2]/2 - 16ln(4 - x) + k
Let k + 32 = c:
--> ∫ x²/(4-x) dx = c - 8x - [(4 - x)^2]/2 - 16ln(4 - x)

Nima
0
X
new posts
Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• Bournemouth University
Midwifery Open Day at Portsmouth Campus Undergraduate
Wed, 16 Oct '19
• Teesside University
Wed, 16 Oct '19
• University of the Arts London
London College of Fashion – Cordwainers Footwear and Bags & Accessories Undergraduate
Wed, 16 Oct '19

### Poll

Join the discussion

#### How has the start of this academic year been for you?

Loving it - gonna be a great year (113)
17.82%
It's just nice to be back! (172)
27.13%
Not great so far... (227)
35.8%
I want to drop out! (122)
19.24%