# Help with integration

1. Jun 27, 2006

### prace

Hello,

I have a question regarding some simple integrals. For the life of my I cannot seem to get the right answer. I know the solution to the problem, but my answer seems to missing an ever so important (1/2) from the equation. Here is what I have:

The probelm is this, ∫tan^(-1)x dx.
[This is the indefinate integral of the inverse tangent of x]

When I work the problem out, my answer comes out to be:

xtan^(-1)x - ln|1+x²|+C

However, the answer that my professor and calculator give me is

xtan^(-1)x - (1/2)ln|1+x²|+C

Any thoughts of where this (1/2) is coming from?

To come to this answer, I first used integration by parts (uv-∫vdu) and used tan^(-1)x as u and (1/(1+x²)) for du. I then took x^0 for dv and x as v.

The next substitution I made was to set x = tan(θ) which made dx = sec²(θ) dθ.

To find θ after I completed the integration, I made a triangle and found that sec(θ) was 1+x². Substituting that back into the equation gives me my final answer of:

xtan^(-1)x - ln|1+x²|+C.

I hope that explaination made sense. I can explain in more depth if needed.

Thank you,

Peter

2. Jun 27, 2006

### 0rthodontist

You don't have to do nearly so much work in the second integral--just substitute u = x^2 + 1. Most likely your error was in forgetting to introduce a square root sign when you determine sec(θ), but I'm not sure where that comes in at all since when I do it with your substitution I have to determine cos(θ) instead.

Last edited: Jun 27, 2006
3. Jun 27, 2006

### prace

Hi,

Thanks for the reply. I tried to substitute u = x^2 + 1, but then du will then = 2x dx no? The second integral does not have another x to use. I don't know if I am saying that clearly or not. I guess what I am trying to say is that if the integral is to evaluate 1/(1+x^2) dx, then you cannot choose to substitute u as 1 + x^2 because du will not make sence. That is why you have to solve the eqation with trig substitution.

If I am totally off here, please let me know. This is just my thinking and if you know a way so that I do not have to use trig sub and just u sub then that would be awesome.

As for the second part of your response, I don't see where I should have a square root sign. Is there any way I could ask you to elaborate a little more?

Thank you so much for your help so far, I really appreciate everything.

Peter

4. Jun 27, 2006

### arildno

No, Prace.
You have clearly misunderstood how to do this, so I'll go through the details:
$$\int\tan^{-1}(x)dx=x\tan^{-1}(x)-\int\frac{xdx}{1+x^{2}}$$
The substitution $u=1+x^{2}\to{du}=2xdx$ leads to:
$$\int\frac{xdx}{1+x^{2}}=\frac{1}{2}\int\frac{du}{u}=\frac{ln|u|}{2}+C=\frac{ln(1+x^{2})}{2}+C$$

5. Jun 27, 2006

### d_leet

You definitely do NOT have to usea trig substitution.

If u = 1 + x2

then
du = 2x*dx
so it follows that
du/2 = x*dx which is what you should have in that integral and so you can substitute du/2 in.

6. Jun 27, 2006

### prace

OH MAN!! Wow, I can't believe I overlooked that. It completely makes sense now. Arrr... I hate errors like that where it is not misunderstanding of the technique or the problem, but a careless error that could have totally been avoided. Thank you so much for everyone's help and sorry if I wasted peoples time with careless errors.