With a view to expressing the problem in the form: :$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$ let: {{begin-eqn}} {{eqn | l = u       | r = \frac 1 {\paren {a x^2 + b x + c}^n}       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d u} {\d x}       | r = \frac {-\paren {2 a x + b} n} {\paren {a x^2 + b x + c}^{n + 1} }       | c = Derivative of Power and Chain Rule for Derivatives }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\d v} {\d x}       | r = x^m       | c =  }} {{eqn | ll= \leadsto       | l = v       | r = \frac {x^{m + 1} } {m + 1}       | c = Primitive of Power }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {x^m \rd x} {\paren {a x^2 + b x + c}^n}       | r = \int \frac 1 {\paren {a x^2 + b x + c}^n} x^m \rd x       | c =  }} {{eqn | r = \frac 1 {\paren {a x^2 + b x + c}^n} \frac {x^{m + 1} } {m + 1} - \int \frac {x^{m + 1} } {m + 1} \frac {-\paren {2 a x + b} n} {\paren {a x^2 + b x + c}^{n + 1} } \rd x       | c = Integration by Parts }} {{eqn | n = 1       | r = \frac {x^{m + 1} } {\paren {m + 1} \paren {a x^2 + b x + c}^n} + \frac n {m + 1} \int \frac {\paren {2 a x + b} x^{m + 1} } {\paren {a x^2 + b x + c}^{n + 1} } \rd x       | c = simplifying }} {{end-eqn}} Meanwhile: {{begin-eqn}} {{eqn | o =        | r = \int \frac {x^m \rd x} {\paren {a x^2 + b x + c}^n}       | c =  }} {{eqn | r = \frac 1 a \int \frac {x^{m - 2} a x^2 \rd x} {\paren {a x^2 + b x + c}^n}       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac 1 a \int \frac {x^{m - 2} \paren {a x^2 + b x + c - b x - c} \rd x} {\paren {a x^2 + b x + c}^n}       | c =  }} {{eqn | r = \frac 1 a \int \frac {x^{m - 2} \rd x} {\paren {a x^2 + b x + c}^{n - 1} }       | c = Linear Combination of Integrals }} {{eqn | o =       | ro= -       | r = \frac b a \int \frac {x^{m - 1} \rd x} {\paren {a x^2 + b x + c}^n} - \frac c a \int \frac {x^{m - 2} \rd x} {\paren {a x^2 + b x + c}^n}       | c = and simplification }} {{eqn | r = \frac 1 a \paren {\frac {x^{m - 1} } {\paren {m - 1} \paren {a x^2 + b x + c}^{n - 1} } + \frac {n - 1} {m - 1} \int \frac {\paren {2 a x + b} x^{m - 1} } {\paren {a x^2 + b x + c}^n} \rd x}       | c = from $(1)$ }} {{eqn | o =       | ro= -       | r = \frac b a \int \frac {x^{m - 1} \rd x} {\paren {a x^2 + b x + c}^n} - \frac c a \int \frac {x^{m - 2} \rd x} {\paren {a x^2 + b x + c}^n}       | c = and simplification }} {{eqn | r = \frac {x^{m - 1} } {a \paren {m - 1} \paren {a x^2 + b x + c}^{n - 1} } + \frac {2 \paren {n - 1} } {m - 1} \int \frac {x^m} {\paren {a x^2 + b x + c}^n} \rd x       | c = multiplying out }} {{eqn | o =       | ro= +       | r = \frac {b \paren {n - 1} } {a \paren {m - 1} } \int \frac {x^{m - 1} } {\paren {a x^2 + b x + c}^n} \rd x       | c =  }} {{eqn | o =       | ro= -       | r = \frac b a \int \frac {x^{m - 1} \rd x} {\paren {a x^2 + b x + c}^n} - \frac c a \int \frac {x^{m - 2} \rd x} {\paren {a x^2 + b x + c}^n}       | c =  }} {{eqn | ll= \leadsto       | o =        | r = \paren {1 - \frac {2 \paren {n - 1} } {m - 1} } \int \frac {x^m \rd x} {\paren {a x^2 + b x + c}^n}       | c = rearranging }} {{eqn | r = \frac {x^{m - 1} } {a \paren {m - 1} \paren {a x^2 + b x + c}^{n - 1} }       | c =  }} {{eqn | o =       | ro= +       | r = \paren {\frac {b \paren {n - 1} } {a \paren {m - 1} } - \frac b a} \int \frac {x^{m - 1} } {\paren {a x^2 + b x + c}^n} \rd x       | c =  }} {{eqn | o =       | ro= -       | r = \frac c a \int \frac {x^{m - 2} \rd x} {\paren {a x^2 + b x + c}^n}       | c =  }} {{eqn | ll= \leadsto       | o =        | r = \int \frac {x^m \rd x} {\paren {a x^2 + b x + c}^n}       | c =  }} {{eqn | r = \frac {x^{m - 1} } {\paren {2 n - m - 1} a \paren {a x^2 + b x + c}^{n - 1} }       | c =  }} {{eqn | o =       | ro= +       | r = \frac {\paren {m - 1} c} {\paren {2 n - m - 1} a} \int \frac {x^{m - 2} \rd x} {\paren {a x^2 + b x + c}^n}       | c =  }} {{eqn | o =       | ro= -       | r = \frac {\paren {n - m} b} {\paren {2 n - m - 1} a} \int \frac {x^{m - 1} \rd x} {\paren {a x^2 + b x + c}^n}       | c = simplification }} {{end-eqn}} {{qed}}