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

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Theorem

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

Then:

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


Proof

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

$\blacksquare$


Sources

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