Primitive of Reciprocal of x by Cube of Root of a x squared plus b x plus c

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Theorem

Let $a \in \R_{\ne 0}$.

Then:

$\ds \int \frac {\d x} {x \paren {\sqrt {a x^2 + b x + c} }^3} = \frac 1 {c \sqrt {a x^2 + b x + c} } + \frac 1 c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } - \frac b {2 c} \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3}$


Proof

\(\ds \) \(\) \(\ds \int \frac {\d x} {x \paren {\sqrt {a x^2 + b x + c} }^3}\)
\(\ds \) \(=\) \(\ds \int \frac {c \rd x} {c x \paren {\sqrt {a x^2 + b x + c} }^3}\) multiplying top and bottom by $c$
\(\ds \) \(=\) \(\ds \int \frac {\paren {a x^2 + b x + c - a x^2 - b x} \rd x} {c x \paren {\sqrt {a x^2 + b x + c} }^3}\)
\(\ds \) \(=\) \(\ds \frac 1 c \int \frac {\paren {a x^2 + b x + c} \rd x} {x \paren {\sqrt {a x^2 + b x + c} }^3} - \frac a c \int \frac {x^2 \rd x} {x \paren {\sqrt {a x^2 + b x + c} }^3} - \frac b c \int \frac {x \rd x} {x \paren {\sqrt {a x^2 + b x + c} }^3}\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \frac 1 c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } - \frac a c \int \frac {x \rd x} {\paren {\sqrt {a x^2 + b x + c} }^3} - \frac b c \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3}\) simplifying
\(\ds \) \(=\) \(\ds \frac 1 c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }\)
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \frac a c \paren {\frac {2 \paren {b x + 2 c} } {\paren {b^2 - 4 a c} \sqrt {a x^2 + b x + c} } }\) Primitive of $\dfrac x {\paren {\sqrt {a x^2 + b x + c} }^3}$
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \frac b c \paren {\frac {2 \paren {2 a x + b} } {\paren {4 a c - b^2} \sqrt {a x^2 + b x + c} } }\) Primitive of $\dfrac 1 {\paren {\sqrt {a x^2 + b x + c} }^3}$
\(\ds \) \(=\) \(\ds \frac 1 c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } + \frac 1 c \paren {\frac {\paren {4 a b x + 2 b^2} - \paren {2 a b x + 4 a c} } {\paren {b^2 - 4 a c} \sqrt {a x^2 + b x + c} } }\) placing over a common denominator
\(\ds \) \(=\) \(\ds \frac 1 c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } + \frac 1 c \paren {\frac {2 a b x + 2 b^2 - 4 a c} {\paren {b^2 - 4 a c} \sqrt {a x^2 + b x + c} } }\) simplifying
\(\ds \) \(=\) \(\ds \frac 1 c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } + \frac 1 c \paren {\frac {b^2 - 4 a c} {\paren {b^2 - 4 a c} \sqrt {a x^2 + b x + c} } } + \frac 1 c \paren {\frac {2 a b x + b^2} {\paren {b^2 - 4 a c} \sqrt {a x^2 + b x + c} } }\) splitting
\(\ds \) \(=\) \(\ds \frac 1 {c \sqrt {a x^2 + b x + c} } + \frac 1 c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } - \frac b {2 c} \paren {\frac {2 \paren {2 a x + b} } {\paren {4 a c - b^2} \sqrt {a x^2 + b x + c} } }\) rearranging and simplifying
\(\ds \) \(=\) \(\ds \frac 1 {c \sqrt {a x^2 + b x + c} } + \frac 1 c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } - \frac b {2 c} \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3}\) Primitive of $\dfrac 1 {\paren {\sqrt {a x^2 + b x + c} }^3}$

$\blacksquare$


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