Please show me how to integrate the following.
[(x+2)/(x+1)](dx)
[(x+2)/(x+1)](dx)
ok, integral of [(x+2)/(x+1)](dx)...
Intergration by parts is ∫udv/dx dx= uv - ∫v du/dx dx
so uv = ∫udv/dx dx+∫v du/dx dx
du/dx = x+2
dv/dx = (x+1)^-1
uv= (x^2+2x)((x+1)^-1) + (x+2)(log(x+1))
that sounds really wrong
i am so bad at this stuff now
Intergration by parts is ∫udv/dx dx= uv - ∫v du/dx dx
so uv = ∫udv/dx dx+∫v du/dx dx
du/dx = x+2
dv/dx = (x+1)^-1
uv= (x^2+2x)((x+1)^-1) + (x+2)(log(x+1))
that sounds really wrong
i am so bad at this stuff now
Last edited by mkxiii (2010-03-03 10:19:43)
mkxiii wrote:
integrate
im a spastic
Last edited by Doctor Strangelove (2010-03-03 10:03:20)
tried again but i think i failedDoctor Strangelove wrote:
mkxiii wrote:
integrate
im a spastic
i saw this a month ago in school
i already forgot how to do this
i already forgot how to do this
i havent had to intergrate anything that like this for about 2 years so i cant do it for shit now
apparently it should be x+log(x+1)
apparently it should be x+log(x+1)
Last edited by mkxiii (2010-03-03 10:21:33)
ok let me try
this is undefined integral, right?
[(x+2)/(x+1)](dx)
= ∫[(x+2).(x+1)^-1].dx
then i'd do the following, but I'm pretty certain it's wrong
NVM, so wrong
this is undefined integral, right?
[(x+2)/(x+1)](dx)
= ∫[(x+2).(x+1)^-1].dx
then i'd do the following, but I'm pretty certain it's wrong
NVM, so wrong
integrating by parts should work just fine
cba to work it out myself, check wolfram
cba to work it out myself, check wolfram
Yeah, I'm not supposed to need to integrate by parts for this one.
Factor out (x+2), integrate the 1/(x+1), result (x+2)ln(x+1) + C?
I dunno, i fail at calc. and i'm in calc 2.
http://www.wolframalpha.com/input/?i=integrate+[%28x%2B2%29%2F%28x%2B1%29]%28dx%29+
I dunno, i fail at calc. and i'm in calc 2.
http://www.wolframalpha.com/input/?i=integrate+[%28x%2B2%29%2F%28x%2B1%29]%28dx%29+
Last edited by Bevo (2010-03-03 10:47:32)
quotient rule? =>∫dv/u=v/u+∫v/(u^2)Doctor Strangelove wrote:
Yeah, I'm not supposed to need to integrate by parts for this one.
Integrate by parts with:
u = x+2 and dv/dx = 1/x+1 I would have thought.
Also:
u = x+2 and dv/dx = 1/x+1 I would have thought.
Also:
...+Cmkxiii wrote:
ok, integral of [(x+2)/(x+1)](dx)...
Intergration by parts is ∫udv/dx dx= uv - ∫v du/dx dx
so uv = ∫udv/dx dx+∫v du/dx dx
du/dx = x+2
dv/dx = (x+1)^-1
uv= (x^2+2x)((x+1)^-1) + (x+2)(log(x+1))
that sounds really wrong
i am so bad at this stuff now
do long division to make it two integrals
Any reason this can't go in the homework help thread?
Any reason this can't go in the homework help thread?
Because it was for a quiz that I needed help with.Flaming_Maniac wrote:
do long division to make it two integrals
Any reason this can't go in the homework help thread?
One that ended 19 minutes ago.
And also, it was one of those things were it starts off all like
(x+1)/(x+1)(x+3)
and then you gotta make it all like
[x+1] = (A(x+3)+B(x+1))/(x+3)(x+1)
And then factor out A and B and shit and then solve for A and B and you get two different integrals that you could solve for. However I couldn't solve for the integrals I got, because this shit sucks.
you needed to divide again for one of the integrals
I guess this is a bit late but...
x + 2 = x + 1 + 1
=> (x + 2) / (x + 1) = 1 + 1 / (x + 1)
Which you can integrate without integration by parts.
You'd get
x + ln(x + 1)
x + 2 = x + 1 + 1
=> (x + 2) / (x + 1) = 1 + 1 / (x + 1)
Which you can integrate without integration by parts.
You'd get
x + ln(x + 1)
Last edited by Arc (2010-03-03 14:16:59)