Indefinite Integral - By parts works right?

[RoganSarine]

Nevermind, it's late and I realized why it doesn't work because I forgot to take into consideration that the denominator is (1/polynomial)

Anyone care to explain to me how to do it the proper way?



1. Question 1

\int (x+2)/(x²+x+1) dx

The only reason I ask is because my teacher showed us another way which basically involves breaking the numerator up, and completing the square in the bottom... which I really don't care to learn. The answer DOES look nicer, but I just want to make sure my way is correct also.

The Attempt at a Solution

uv - \intvdu

u = x+2
du = 1

dv = x²+x+1
v = (1/3)x^3 + (1/2)x^2 + x

(x+2)[(1/3)x^3 + (1/2)x^2 + x] - \int [ (1/3)x^3 + (1/2)x^2 + x ] (1)

(x+2)[(1/3)x^3 + (1/2)x^2 + x] - [(1/4)x^4 -(1/6)x^3 + (1/2)x^2] + C

As I said... Not really pretty by any means, but isn't that technically also correct?[/s]

 

[tiny-tim]

Anyone care to explain to me how to do it the proper way?

Hi RoganSarine!

(x+2)/(x²+x+1)

= (x+2)/((x + 1/2)² + 3/4)

= (x + 1/2)/((x + 1/2)² + 3/4) + (3/2)/((x + 1/2)² + 3/4) …

now integrate (you'll need a different method for each part )

 

[RoganSarine]

I can see where to get most of that equation but that last part:

(x+2)/(x²+x+1)

(x+1/2+3/2)/(x²+1/2+1/4-1/4)+1)
(x+1/2+3/2)/((x+1/2)²+1-1/4)
(x+1/2+3/2)/((x+1/2)²+3/4)


But...Oh nevermind.

Wow, ha ha... I've never had to break up an equation like that before (like, breaking up the numerator also, seriously?). I'll give it a shot.