been erasing for an hour now-i give up :( someone help
integral (x^2+5)/(x^3-x^2+x+3)
integral (x^2+5)/(x^3-x^2+x+3)
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For the denominator... if you remember from Pre-Calc, you can figure out that (x+1) is a factor of (x^3-x^2+x+3).
So, if you divide the denominator by (x+1) using long or synthetic division, you can figure out the the denominator factors out into (x+1)(x^2-2x+3).
Unfortunately, you can't factor (x^2-2x+3) anymore.
Next:
(x^2+5)/ (x+1)(x^2-2x+3) = A/(x+1) + (Bx+C)/ (x^2-2x+3)
Solve for A, B and C. You'll find that A= 1, B=0, and C= 2
Plug it back in to the integration thingie:
integration of ((1)/(x+1))+((2)/(x^2-2x+3)) dx
You can integrate the first part easily to get ln(x+1)
the second part, you are left with: integration of 2/(x^2-2x+3)
look at the denominator and just follow the next steps:
x^2-2x+3 is equal to x^2-2x+1+2... if you factor just the first three parts of that, you will get: (x-1)^2+2
now you have the integration of 2/((x-1)^2+2) dx
You should recognize that this looks like an arctan function (1/ u^2+1)
use the arctan function rule to solve
So, if you divide the denominator by (x+1) using long or synthetic division, you can figure out the the denominator factors out into (x+1)(x^2-2x+3).
Unfortunately, you can't factor (x^2-2x+3) anymore.
Next:
(x^2+5)/ (x+1)(x^2-2x+3) = A/(x+1) + (Bx+C)/ (x^2-2x+3)
Solve for A, B and C. You'll find that A= 1, B=0, and C= 2
Plug it back in to the integration thingie:
integration of ((1)/(x+1))+((2)/(x^2-2x+3)) dx
You can integrate the first part easily to get ln(x+1)
the second part, you are left with: integration of 2/(x^2-2x+3)
look at the denominator and just follow the next steps:
x^2-2x+3 is equal to x^2-2x+1+2... if you factor just the first three parts of that, you will get: (x-1)^2+2
now you have the integration of 2/((x-1)^2+2) dx
You should recognize that this looks like an arctan function (1/ u^2+1)
use the arctan function rule to solve