## Integrating a complicated polynomial

1. The problem statement, all variables and given/known data
Hi
An integration question:
t= 1/(1+r2)

Can you please show me how to integrate with respect to r. Thank you.

3. The attempt at a solution
∴t = (1+r2)-1
I then tried substituting u = 1+r^2 but that didn't work!
Is there a trick with this?
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 Recognitions: Gold Member Well I'll hint and say it has to do with an inverse trig function.

Recognitions:
Homework Help
Just so I understand, you're trying to find the integral of this?
$$\int \frac{1}{1+r^2}\,dr$$
BTW, what you have after the t is not a polynomial, but a rational expression.
 Quote by Mathpower 3. The attempt at a solution ∴t = (1+r2)-1 I then tried substituting u = 1+r^2 but that didn't work! Is there a trick with this?
Yes, it's called trigonometric substitution. Try r = tan u.

## Integrating a complicated polynomial

eumyang...thank you...yes thats what i meant.
 NO idea! I tried r= tan u 1/(1+(tan u)^2)... Too me that just seems even more worse! Then (LET: < be integral sign/ integrand) $= <1/(1+u) dr = <(1/(1+u)) (secu)^2 du = <((secu)^2/(1+u)) du$ Stuck... and my latex is not working as well? Have I typed the wrong code?

Recognitions:
Homework Help
 Quote by Mathpower NO idea! I tried r= tan u 1/(1+(tan u)^2)... Too me that just seems even more worse! Then (LET: < be integral sign/ integrand) $= <1/(1+u) dr = <(1/(1+u)) (secu)^2 du = <((secu)^2/(1+u)) du$ Stuck...
No no, put the two attempts together:
$$\int \frac{1}{1+r^2}\,dr = \int \frac{1}{1+\tan^2 u}\,\sec^2 u \, du$$
You know your trig identities, right?

EDIT: Regarding LaTeX, that's because you're using the wrong tags. It should be tex or itex in the brackets, not latex.
 Delete this. See below
 Oh wait is the answer: u because 1+tan^2 u = sec^2 u then $$= ∫1 du =u + C$$ =tan(r)^-1 + C Thank you for your help.

Recognitions:
Homework Help
 Quote by Mathpower oh wait is the answer u because 1+tan^2 u = sec^2 u then <1
If you mean
$$\int 1$$
...then you're missing something.
 Quote by Mathpower =u =tan(r)^-1 + C
Do you mean
$\tan r^{-1} + C$
or
$\tan^{-1} r + C$?
 Opps... I meant $$tan-1 r+C$$ Thank you so much for your help. Sorry for being lazy.

 Tags mathpower, question 4

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