First: {{begin-eqn}} {{eqn | o =        | r = \int \frac {\d x} {x^2 \paren {a x^2 + b x + c}^2}       | c =  }} {{eqn | r = \int \frac {c \rd x} {c x^2 \paren {a x^2 + b x + c}^2}       | c = multiplying top and bottom by $c$ }} {{eqn | r = \frac 1 c \int \frac {c \rd x} {x^2 \paren {a x^2 + b x + c}^2}       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac 1 c \int \frac {a x^2 + b x + c - a x^2 - b x} {x^2 \paren {a x^2 + b x + c}^2} \rd x       | c = adding and subtracting $a x^2 + b x$ }} {{eqn | r = \frac 1 c \int \frac {\paren {a x^2 + b x + c} \rd x} {x^2 \paren {a x^2 + b x + c}^2} - \frac a c \int \frac {x^2 \rd x} {x^2 \paren {a x^2 + b x + c}^2}       | c = Linear Combination of Integrals }} {{eqn | o =        | ro= -       | r = \frac b c \int \frac {x \rd x} {x^2 \paren {a x^2 + b x + c}^2}       | c =  }} {{eqn | n = 1       | r = \frac 1 c \int \frac {\d x} {x^2 \paren {a x^2 + b x + c} } - \frac a c \int \frac {\d x} {\paren {a x^2 + b x + c}^2}       | c = simplification }} {{eqn | o =        | ro= -       | r = \frac b c \int \frac {\d x} {x \paren {a x^2 + b x + c}^2}       | c =  }} {{end-eqn}} Next, with a view to obtaining an expression 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} }       | c =  }} {{eqn | r = \paren {a x^2 + b x + c}^{-1}       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d u} {\d x}       | r = -\paren {2 a x + b} \paren {a x^2 + b x + c}^{-2}       | c = Chain Rule for Derivatives and Derivative of Power }} {{eqn | r = \frac {-\paren {2 a x + b} } {\paren {a x^2 + b x + c}^2}       | c = simplifying }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\d v} {\d x}       | r = \frac 1 {x^2}       | c =  }} {{eqn | ll= \leadsto       | l = v       | r = \frac {-1} x       | c = Primitive of Power }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {x^2 \paren {a x^2 + b x + c} }       | r = \int \frac 1 {\paren {a x^2 + b x + c} } \frac 1 {x^2} \rd x       | c =  }} {{eqn | r = \frac 1 {\paren {a x^2 + b x + c} } \frac {-1} x - \int \frac {-1} x \frac {-\paren {2 a x + b} } {\paren {a x^2 + b x + c}^2} \rd x       | c = Integration by Parts }} {{eqn | n = 2       | r = \frac {-1} {x \paren {a x^2 + b x + c} } - 2 a \int \frac {\d x} {\paren {a x^2 + b x + c}^2}       | c = simplification }} {{eqn | o =        | ro= -       | r = b \int \frac {\d x} {x \paren {a x^2 + b x + c}^2}       | c = Linear Combination of Integrals }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | o =        | r = \int \frac {\d x} {x^2 \paren {a x^2 + b x + c}^2}       | c =  }} {{eqn | r = \frac 1 c \int \frac {\d x} {x^2 \paren {a x^2 + b x + c} } - \frac a c \int \frac {\d x} {\paren {a x^2 + b x + c}^2} - \frac b c \int \frac {\d x} {x \paren {a x^2 + b x + c}^2}       | c = from $(1)$ }} {{eqn | r = \frac 1 c \paren {\frac {-1} {x \paren {a x^2 + b x + c} } - 2 a \int \frac {\d x} {\paren {a x^2 + b x + c}^2} - b \int \frac {\d x} {x \paren {a x^2 + b x + c}^2} }       | c = from $(2)$ }} {{eqn | o =        | ro= -       | r = \frac a c \int \frac {\d x} {\paren {a x^2 + b x + c}^2} - \frac b c \int \frac {\d x} {x \paren {a x^2 + b x + c}^2}       | c =  }} {{eqn | r = \frac {-1} {c x \paren {a x^2 + b x + c} } - \frac {3 a} c \int \frac {\d x} {\paren {a x^2 + b x + c}^2} - \frac {2 b} c \int \frac {\d x} {x \paren {a x^2 + b x + c}^2}       | c = gathering terms }} {{end-eqn}} {{qed}}