# Indefinate Integral 1/(x^2 +1)

1. Jan 7, 2014

### Calu

1. The problem statement, all variables and given/known data

This is the very last piece of an integration using partial fractions question. I have to integrate 1/x2+1.

2. Relevant equations

Tan2x + 1 = Sec2x

3. The attempt at a solution

Substitute tanu=x:

1/tan2u + 1 = 1/sec2u = cos2u

which leaves me with int [ cos2u ] .dx I'm not sure what to do from here. Or if what I've done is correct.

2. Jan 7, 2014

### gopher_p

Do you know a function whose derivative is $\frac{1}{x^2+1}$?

3. Jan 7, 2014

### D H

Staff Emeritus
You haven't complete the u-substitution. You need to express dx as (something)*du. You used the u-substitution tan(u)=x, so what does that mean dx is?

4. Jan 7, 2014

### Calu

I'm not sure, that's the bit I'm having difficulty with.

5. Jan 7, 2014

### D H

Staff Emeritus
What is $\frac {d}{du} (\tan u)$ ?

6. Jan 7, 2014

### Calu

I think I have it. I have

d/dx(tanu)=d/dx(x)
du/dx(sec2u)=1
du/dx=1/sec2u
dx=sec2u .du

but I believe this to be incorrect, as that leaves me with

∫cos2x.sec2x.du = ∫1 .du = u.

7. Jan 7, 2014

### Calu

I believe that is sec2u, but I don't see how it relates to dx.

Last edited: Jan 7, 2014
8. Jan 7, 2014

### LCKurtz

Aside from the missing constant of integration, that is correct. You started with $x=\tan u$. What do you get if you solve for $u$ to get your answer in terms of $x$?

9. Jan 7, 2014

### Calu

Ahh, I see now. arctanx=u. So ∫1/x2 + 1 .dx = arctanx +c. Thanks.

10. Jan 7, 2014

### Staff: Mentor

You need to use parentheses when there are multiple terms in the top or bottom of a fraction.
Write this as 1/(x2 + 1). What you wrote is (1/x2) + 1, which isn't what you meant.

Both examples above should be 1/(tan2(u) + 1).