Question

How to solve integral of ((x^2+5)/(x^3-x^2+x+3))dx?

Answer 1

\[\int\limits \frac{ x^2+5 }{ x^3-x^2+x+3 }dx\]
sorry, i made a typo in the original question.

Answer 2

more specifically, how would you split that into partial fractions?

Answer 3

i think i'm just a little confused on how to split up the denominator.

Answer 4

i think by some miracle the denominator factors

Answer 5

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

Answer 6

how did you get that from the denominator? cuz i got (x^2-x+3)(x-1)

Answer 7

whoops, i meant (x^2+x+3)(x-1)

Answer 8

easy, i cheated
http://www.wolframalpha.com/input/?i=+ \frac{+x^2%2B5+}{+x^3-x^2%2Bx%2B3+}

Answer 9

it can't be \(x-1\) because1 is not a zero of the denominator, \(-1\) is

Answer 10

so it must be \(x+1\) as one factor
you can get the other factor by division

Answer 11

cuz i did x^2(x-1)+x+3 which becomes (x^2+x+3)(x-1)

Answer 12

can't be

Answer 13

at the risk of repeating myself, \(-1\) is a zero of the denominator so it MUST factor as \((x+1)(something)\)

Answer 14

ASDFGHJKL./;"
i think i just figured it out. i think.

Answer 15

good!

Answer 16

cheat and use wolfram
the problem was cooked so that the numerators are integers when you get partial fractions

Answer 17

cooked?

Answer 18

i got the answer anyways.

Answer 19

@jennychan12
Was\[\log (x+1)+\sqrt{2} \tan ^{-1}\left(\frac{x-1}{\sqrt{2}}\right) \]your answer?

Answer 20

i think so; i don't really remember it cuz i turned in the paper.
thanks tho! @robtobey