# integration..

#### ice_syncer

Integrate f(x) = 1/(n^2 + n + 1) with respect to n... also, can it by splitting it into partial fractions, in terms of its factors? Yeah I know wel'' get comlex rots.. can we still integrate it using that ? well, I cant really split into partials without factoriing it... everytime I get value of denominators as 1 hence getting back the same unsplit function

#### CaptainBlack

MHF Hall of Fame
Integrate f(x) = 1/(n^2 + n + 1) with respect to n... also, can it by splitting it into partial fractions, in terms of its factors? Yeah I know wel'' get comlex rots.. can we still integrate it using that ? well, I cant really split into partials without factoriing it... everytime I get value of denominators as 1 hence getting back the same unsplit function
Complete the square in the denominator and after a change of variable you have a standard integral.

(complex partial fractions will also work).

(I will move this to the calculus area as soon as it is visible to the naked eye)

CB

#### Ackbeet

MHF Hall of Honor
I would complete the square in the denominator, and then maybe use trig substitution.

#### ice_syncer

1/((n+0.5)^2 +3/4) .. then what?

#### Ackbeet

MHF Hall of Honor
How about a simple u substitution to continue?

#### ice_syncer

n+1/2 == sec(theta)

#### Ackbeet

MHF Hall of Honor
Well, you can skip steps if you like. That's going too fast for the likes of me. I would just substitute $$\displaystyle v=n+0.5$$ first, since you don't gain any complexity.

I do trig substitutions all the same way. First step: draw a triangle. Next step, assign names to sides. If the squares in the integrand are added, make the hypotenuse equal to the sum of the squares of the other two sides. If the squares have opposite sign, then make the positive one the hypotenuse, and one of the other sides gets to be the negative one. Assign an angle somewhere, and you're off and running. The virtue of doing it this way, is that often the inverse substitution (for indefinite integrals, this is required) is quite apparent straight from the triangle.

:/ hmm ok

#### TheCoffeeMachine

You can avoid substitution!

Complete the square -- $$\displaystyle \displaystyle\int\frac{1}{n^2+n+1}\;{dn} = \int\frac{1}{\left(n+\frac{1}{2}\right)^2+\frac{3}{4}}\;{dn} = \int\frac{1}{\left(n+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}\;{dn}$$ -- and then recall that $$\displaystyle \displaystyle\int\dfrac{1}{x^2+a^2}\;{dx} = \dfrac{1}{a}\arctan\dfrac{x}{a}+C.$$

#### Ackbeet

MHF Hall of Honor
Well, to each his own. I would rather remember trig substitution than have to remember one particular integral.