Editing Partial fraction decomposition (section)

===Example 6 (integral)===
Suppose we have the indefinite [[integral]]:

<math display="block">\int \frac{x^4+x^3+x^2+1}{x^2+x-2} \,dx</math>

Before performing decomposition, it is obvious we must perform polynomial long division and [[Factorization|factor]] the denominator. Doing this would result in:

<math display="block">\int \left(x^2 + 3 + \frac{-3x+7}{(x+2)(x-1)}\right) dx</math>

Upon this, we may now perform partial fraction decomposition.

<math display="block">\int \left(x^2+3+ \frac{-3x+7}{(x+2)(x-1)}\right) dx = \int \left(x^2+3+ \frac{A}{(x+2)}+\frac{B}{(x-1)}\right) dx</math>
so:
<math display="block">A(x-1)+B(x+2)=-3x+7</math>.
Upon substituting our values, in this case, where x=1 to solve for B and x=-2 to solve for A, we will result in:

<math display="block">A=\frac{-13}{3} \ , B=\frac{4}{3} </math>

Plugging all of this back into our integral allows us to find the answer:

<math display="block">\int \left(x^2+3+ \frac{-13/3}{(x+2)}+\frac{4/3}{(x-1)}\right) \,dx = \frac{x^3}{3} \ + 3x-\frac{13}{3} \ln(|x+2|)+\frac{4}{3} \ln(|x-1|)+C </math>