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Quick integration problem :)

How would I integrate

x/(1+x) dx

I used U substitution and used u = 1+x

So du = dx

X = 1-u

So I had (1-u)/u which gives

(1/u) -1

Integrating that I got Ln(u) - u and substituting back in gave ln(1+x) -(1+x) which is ln(1+x) - 1 - x

But, the answer in the AQA book and on my TI89 gives x - ln(1+x)

Hope somebody can help.

Thanks very much,
Ross
Reply 1
Avatar for cj_134
cj_134
9
no x = u-1
Reply 2
Avatar for jit987
jit987
13
x/(1+x) can be written as: 1 - 1/(1+x).
Reply 3
Avatar for dknt
dknt
12
Original post by rosschambers1987
How would I integrate

x/(1+x) dx

I used U substitution and used u = 1+x

So du = dx

X = 1-u

So I had (1-u)/u which gives

(1/u) -1

Integrating that I got Ln(u) - u and substituting back in gave ln(1+x) -(1+x) which is ln(1+x) - 1 - x

But, the answer in the AQA book and on my TI89 gives x - ln(1+x)

Hope somebody can help.

Thanks very much,
Ross


You don't have to use a substitution. Note that for x/(1+x), the numerator is of equal degree to the denominator, as such, you can do long division. With so few terms that should be very quick. You will then be left with some terms which you can easily integrate by inspection.
Reply 4
Avatar for nuodai
nuodai
17
Original post by rosschambers1987
How would I integrate

x/(1+x) dx

I used U substitution and used u = 1+x

So du = dx

X = 1-u

So I had (1-u)/u which gives

(1/u) -1

Integrating that I got Ln(u) - u and substituting back in gave ln(1+x) -(1+x) which is ln(1+x) - 1 - x

But, the answer in the AQA book and on my TI89 gives x - ln(1+x)

Hope somebody can help.

Thanks very much,
Ross


You should have put x=u−1x=u-1 instead (you just rearranged u=1+x wrong), which is why your answer is -1 times what it should be. And don't worry about the 1 floating around at the end: the arbitrary constant of integration takes care of that.

However, it's just as easy to just split x1+x\dfrac{x}{1+x} up directly. Notice that x=(1+x)−1x=(1+x)-1, and so x1+x=1−11+x\dfrac{x}{1+x} = 1 - \dfrac{1}{1+x}. If you do this then you don't need to use a substitution.
Reply 5
Avatar for deezykhara
deezykhara
use algebraic division on x/(1+x) to get 1- (1/(1+x)) and then integrate and you get the answer given
Reply 6
Avatar for DFranklin
DFranklin
18
@nuodai: Don't you "need" a substitution to integrate 11+x\frac{1}{1+x}?
Original post by DFranklin
@nuodai: Don't you "need" a substitution to integrate 11+x\frac{1}{1+x}?

It should go straight to a log. I can see where you're going with what you said though... Most people doing A-Level maths have learnt the way to deal with integrals that lead to logarithms by inspection rather than substitution. Don't know if that's a bad thing or not, really.
Hi everyone, thanks for posting back to me. Very helpful feedback :smile:

It's a load of questions in an AQA book where I have to choose the best method to integrate so there's a mix of all different techniques. I suppost practice on these will lead me to choose the correct method in future.

Thanks again, most helpful!
Reply 9
Avatar for DFranklin
DFranklin
18
Original post by Farhan.Hanif93
It should go straight to a log. I can see where you're going with what you said though... Most people doing A-Level maths have learnt the way to deal with integrals that lead to logarithms by inspection rather than substitution. Don't know if that's a bad thing or not, really.
I had no idea what people would/wouldn't know at A-level these days.
I only started studying A level maths in Sept last year, doing a full A level, C1, C2, C3, C4, MFP1 and MFP3.

Done 12 and F1 and have my 3,4 and FP3 exams in June.

Cramming! All I've learned is from the internet and some books I have.

Managed a B in my AS and need B in the other 3 to get into Uni!

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