integration using arctan

**collegestudent321** · February 19th 2009, 02:20 PM

could anyone possibly help me with this problem:

integral of: 3/((x^2)+2x+5)???

The only thing I know is that you have to complete the square for the denominator and that gives me:

(x+1)^2 + 4.

I also know that this type of integral uses some variation of arctan, but after this, I'm completely lost.

**Krizalid** · February 19th 2009, 03:17 PM

Next step: make the substitution $x+1=2\tan\varphi.$

**collegestudent321** · February 19th 2009, 03:49 PM

could you explain why that substitution is made? or is it just common knowledge?

**Krizalid** · February 19th 2009, 03:55 PM

The standard arctan form is $\frac1{1+x^2},$ somehow, you just need to turn your integral into that form.

Forget what I said before, I had my head in another place; the substitution it's actually $x+1=2u\implies dx=2\,du$ so $(x+1)^2+4$ becomes $4u^2+4=4(u^2+1)$ and the integral is $\frac12\int\frac{du}{1+u^2}=\frac12\arctan u+k.$

**collegestudent321** · February 19th 2009, 04:04 PM

ok, that all makes sense to me except for the substitution still. I'm sorry if I am bugging you, but I just want to make sure I understand. Why is it 2u and not u?

**Krizalid** · February 19th 2009, 04:09 PM

I'll put another example to see if you get it:

$\int\frac{dx}{x^2+6x+12},$ what's the proper substitution?

**collegestudent321** · February 19th 2009, 04:20 PM

is it: sqrt(3)u = x+3??

**Krizalid** · February 19th 2009, 04:21 PM

Jackpot!!

**collegestudent321** · February 19th 2009, 04:22 PM

Cool, ok I get it now. wow thank you sooo much! you really helped me out.

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