Question

Hey malevolence help me out

Integral of (x^2-4x+7)/(x^3-x^2+x+3)

Answer 1

Factor top and botton, rewrite it as the sum of terms of the form : \[\frac{a}{x+c}\]

Answer 2

Alright.
\[\int\limits\limits \frac{x^2-4x+7}{x^3-x^2+x+3}dx=\int\limits \frac{x^2-4x+7}{(x+1)(x^2-2x+3)}dx\]
Partial fraction decomp:
\[\int\limits \frac{x^2-4x+7}{(x+1)(x^2-2x+3)}dx=\int\limits(\frac{A}{x+1}+\frac{Bx+C}{x^2-2x+3})dx\]
Common denominator gives your numerators to be:
\[x^2-4x+7=Ax^2-2Ax+3A+Bx^2+Cx+Bx+C\]
So:
A+B=1; 3A+C=7; -2A+C+B=-4
Solving this system I get:
B=1-A; C=7-3A
Plugging in to the third equation you get: -2A+7-3A+1-A=-4
A=2.Plugging in the others you get B=1-2=-1 and C=7-3(2)=1.