{{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {\left({a x^2 + b x + c}\right)^{n + \frac 1 2} }       | r = \int \left({ \frac {4 a} {\left({2 a x + b}\right)^2 + 4 a c - b^2} }\right)^{n + \frac 1 2} \ \mathrm d x        | c = Completing the Square }} {{eqn | r = \left({2 \sqrt a}\right)^{2 n + 1} \int \frac {\mathrm d x} {\left({\left({2 a x + b}\right)^2 + 4 a c - b^2}\right)^{n + \frac 1 2} }        | c = Linear Combination of Integrals }} {{end-eqn}} Let: {{begin-eqn}} {{eqn | l = z       | r = \left({2 a x + b}\right)^2 }} {{eqn | ll= \implies       | l = \frac {\mathrm d z} {\mathrm d x}       | r = 2 a \cdot 2 \left({2 a x + b}\right)       | c = Derivative of Power and Chain Rule for Derivatives }} {{eqn | r = 4 a \sqrt z       | c =  }} {{end-eqn}} Also let $q = 4 a c - b^2$. Then: {{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {\left({a x^2 + b x + c}\right)^{n + \frac 1 2} }       | r = \left({2 \sqrt a}\right)^{2 n + 1} \int \frac {\mathrm d z} {4 a \sqrt z \left({z + q}\right)^{n + \frac 1 2} }       | c = Integration by Substitution }} {{eqn | n = 1       | r = \frac {\left({2 \sqrt a}\right)^{2 n + 1} } {4 a} \int \frac {\mathrm d z} {\left({z + q}\right)^{n + \frac 1 2} \sqrt z}       | c = Linear Combination of Integrals }} {{end-eqn}} From Primitive of $\dfrac 1 {\left({p x + q}\right)^n \sqrt{a x + b} }$: :$\displaystyle \int \frac {\mathrm d x} {\left({p x + q}\right)^n \sqrt{a x + b} } = \frac {\sqrt{a x + b} } {\left({n - 1}\right) \left({a q - b p}\right) \left({p x + q}\right)^{n-1} } + \frac {\left({2 n - 3}\right) a} {2 \left({n - 1}\right)  \left({a q - b p}\right)} \int \frac {\mathrm d x} {\left({p x + q}\right)^{n-1} \sqrt{a x + b} }$ Here $p = 1, a = 1, b = 0$ and $n := n + \dfrac 1 2$: {{begin-eqn}} {{eqn | r = \int \frac {\mathrm d x} {\left({a x^2 + b x + c}\right)^{n + \frac 1 2} }       | o =        | c =  }} {{eqn | r = \frac {\left({2 \sqrt a}\right)^{2 n + 1} } {4 a} \int \frac {\mathrm d z} {\left({z + q}\right)^{n + \frac 1 2} \sqrt z}       | c = from $(1)$ }} {{eqn | r = \frac {\left({2 \sqrt a}\right)^{2 n + 1} } {4 a} \left({\frac {\sqrt z} {\left({\left({n + \frac 1 2}\right) - 1}\right) q \left({z + q}\right)^{\left({n + \frac 1 2}\right) - 1} } + \frac {2 \left({n + \frac 1 2}\right) - 3} {2 \left({\left({n + \frac 1 2}\right) - 1}\right) q} \int \frac {\mathrm d z} {\left({z + q}\right)^{\left({n + \frac 1 2}\right) - 1} \sqrt z} }\right)       | c = Primitive of $\dfrac 1 {\left({p x + q}\right)^n \sqrt{a x + b} }$ }} {{eqn | r = \frac {\left({2 \sqrt a}\right)^{2 n + 1} } {4 a} \left({\frac {2 \sqrt z} {\left({2 n - 1}\right) q \left({z + q}\right)^{n - \frac 1 2} } + \frac {2 \left({n - 1}\right)} {\left({2 n - 1}\right) q} \int \frac {\mathrm d z} {\left({z + q}\right)^{n - \frac 1 2} \sqrt z} }\right)       | c = simplifying }} {{eqn | r = \frac {\left({2 \sqrt a}\right)^{2 n + 1} } {4 a} \left({\frac {2 \left({2 a x + b}\right)} {\left({2 n - 1}\right) q \left({\left({2 a x + b}\right)^2 + q}\right)^{n - \frac 1 2} } + \frac {2 \left({n - 1}\right)} {\left({2 n - 1}\right) q} \int \frac {4 a \ \mathrm d x} {\left({\left({2 a x + b}\right)^2 + q}\right)^{n - \frac 1 2} } }\right)       | c = substituting for $z$ }} {{eqn | r = \frac 2 {2 n - 1} \frac {\left({2 a x + b}\right) \left({4 a}\right)^{n - \frac 1 2} } {\left({4 a c - b^2}\right) \left({\left({2 a x + b}\right)^2 + 4 a c - b^2}\right)^{n - \frac 1 2} }       | c = substituting for $q$ }} {{eqn | o =        | ro= +       | r = \frac {8 a \left({n - 1}\right)} {\left({2 n - 1}\right) \left({4 a c - b^2}\right)} \int \frac {\left({4 a}\right)^{n - \frac 1 2} \ \mathrm d x} {\left({\left({2 a x + b}\right)^2 + 4 a c - b^2}\right)^{n - \frac 1 2} }       | c = and simplifying }} {{eqn | r = \frac {2 \left({2 a x + b}\right)} {\left({2 n - 1}\right) \left({4 a c - b^2}\right) \left({a x^2 + b x + c}\right)^{n - \frac 1 2} } + \frac {8 a \left({n - 1}\right)} {\left({2 n - 1}\right) \left({4 a c - b^2}\right)} \int \frac {\mathrm d x} {\left({a x^2 + b x + c}\right)^{n - \frac 1 2} }       | c = Completing the Square }} {{end-eqn}} {{qed}}