Primitive of Root of a x squared plus b x plus c over x squared

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Theorem

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

Then:

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


Proof

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

$\blacksquare$


Sources

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