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

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Theorem

Let $a, b, c \in \R_{\ne 0}$.

Then for $x \in \R$ such that $a x^2 + b x + c > 0$ and $x \ne 0$:

$\ds \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } = \begin {cases} \dfrac {-1} {\sqrt c} \dfrac {\size x} x \ln \size {\dfrac {2 \sqrt c \sqrt {a x^2 + b x + c} + b x + 2 c} x} + C & : c > 0, b^2 - 4 a c > 0 \\ \dfrac {-1} {\sqrt c} \map {\sinh^{-1} } {\dfrac {b x + 2 c} {\size x \sqrt {4 a c - b^2} } } + C & : c > 0, b^2 - 4 a c < 0 \\ \dfrac {-1} {\sqrt c} \dfrac {\size x} x \ln \size {\dfrac {2 c} x + b} + C & : c > 0, b^2 - 4 a c = 0 \\ \dfrac 1 {\sqrt {-c} } \map \arcsin {\dfrac {b x + 2 c} {\size x \sqrt {\size {b^2 - 4 a c} } } } & : c < 0, b^2 - 4 a c \ne 0 \\ \end {cases}$


Proof 1

Lemma

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

where $u := \dfrac 1 x$, according to whether $u > 0$ or $u < 0$.

$\Box$


Let $x > 0$, and so $u > 0$.

Then we have:

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


We consider the two cases where $c > 0$ and $c < 0$.


First we take $c > 0$.

Thus:

\(\ds -\int \frac {\d u} {\sqrt {a + b u + c u^2} }\) \(=\) \(\ds \begin {cases} \dfrac {-1} {\sqrt c} \ln \size {2 \sqrt c \sqrt {a + b u + c u^2} + 2 c u + b} + C & : b^2 - 4 a c > 0 \\ \dfrac {-1} {\sqrt c} \map {\sinh^{-1} } {\dfrac {2 c u + b} {\sqrt {4 a c - b^2} } } + C & : b^2 - 4 a c < 0 \\ \dfrac {-1} {\sqrt c} \ln \size {2 c u + b} + C & : b^2 - 4 a c = 0 \end {cases}\) Primitive of $\dfrac 1 {\sqrt {a x^2 + b x + c} }$
\(\ds \leadsto \ \ \) \(\ds \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }\) \(=\) \(\ds \begin{cases} \dfrac {-1} {\sqrt c} \ln \size {2 \sqrt c \sqrt {a + b \paren {\frac 1 x} + c \paren {\frac 1 x}^2} + 2 c \paren {\frac 1 x} + b} + C & : b^2 - 4 a c > 0 \\ \dfrac {-1} {\sqrt c} \map {\sinh^{-1} } {\dfrac {2 c \paren {\frac 1 x} + b} {\sqrt {4 a c - b^2} } } + C & : b^2 - 4 a c < 0 \\ \dfrac {-1} {\sqrt c} \ln \size {2 c \paren {\frac 1 x} + b} + C & : b^2 - 4 a c = 0 \end{cases}\) substituting for $u$
\(\ds \) \(=\) \(\ds \begin{cases} \dfrac {-1} {\sqrt c} \ln \size {2 \sqrt c \sqrt {\dfrac {a x^2 + b x + c} {x^2} } + \dfrac {b x + 2 c} x} + C & : b^2 - 4 a c > 0 \\ \dfrac {-1} {\sqrt c} \map {\sinh^{-1} } {\dfrac {b x + 2 c} {x \sqrt {4 a c - b^2} } } + C & : b^2 - 4 a c < 0 \\ \dfrac {-1} {\sqrt c} \ln \size {\dfrac {2 c} x + b} + C & : b^2 - 4 a c = 0 \end{cases}\) simplifying
\(\ds \) \(=\) \(\ds \begin{cases} \dfrac {-1} {\sqrt c} \ln \size {\dfrac {2 \sqrt c \sqrt {a x^2 + b x + c} + b x + 2 c} x} + C & : b^2 - 4 a c > 0 \\ \dfrac {-1} {\sqrt c} \map {\sinh^{-1} } {\dfrac {b x + 2 c} {\size x \sqrt {4 a c - b^2} } } + C & : b^2 - 4 a c < 0 \\ \dfrac {-1} {\sqrt c} \ln \size {\dfrac {2 c} x + b} + C & : b^2 - 4 a c = 0 \end{cases}\)

$\Box$


Then we take $c < 0$.

Thus:

\(\ds -\int \frac {\d u} {\sqrt {a + b u + c u^2} }\) \(=\) \(\ds -\dfrac {-1} {\sqrt {-c} } \map \arcsin {\dfrac {2 c u + b} {\sqrt {\size {b^2 - 4 a c} } } }\) Primitive of $\dfrac 1 {\sqrt {a x^2 + b x + c} }$
\(\ds \leadsto \ \ \) \(\ds \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }\) \(=\) \(\ds \dfrac 1 {\sqrt {-c} } \map \arcsin {\dfrac {2 c \paren {\frac 1 x} + b} {\sqrt {\size {b^2 - 4 a c} } } }\) substituting for $u$
\(\ds \) \(=\) \(\ds \dfrac 1 {\sqrt {-c} } \map \arcsin {\dfrac {b x + 2 c} {x \sqrt {\size {b^2 - 4 a c} } } }\) simplifying

$\Box$


Now we consider what happens when $x < 0$, and so $u < 0$.

We have:

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

which leads us to:

\(\ds \int \frac {\d u} {\sqrt {a + b u + c u^2} }\) \(=\) \(\ds \begin {cases} \dfrac 1 {\sqrt c} \ln \size {2 \sqrt c \sqrt {a + b u + c u^2} + 2 c u + b} + C & : b^2 - 4 a c > 0 \\ \dfrac 1 {\sqrt c} \map {\sinh^{-1} } {\dfrac {2 c u + b} {\sqrt{4 a c - b^2} } } + C & : b^2 - 4 a c < 0 \\ \dfrac 1 {\sqrt c} \ln \size {2 c u + b} + C & : b^2 - 4 a c = 0 \end {cases}\) Primitive of $\dfrac 1 {\sqrt {a x^2 + b x + c} }$
\(\ds \leadsto \ \ \) \(\ds \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }\) \(=\) \(\ds \begin{cases} \dfrac 1 {\sqrt c} \ln \size {2 \sqrt c \sqrt {a + b \paren {\frac 1 x} + c \paren {\frac 1 x}^2} + 2 c \paren {\frac 1 x} + b} + C & : b^2 - 4 a c > 0 \\ \dfrac 1 {\sqrt c} \map {\sinh^{-1} } {\dfrac {2 c \paren {\frac 1 x} + b} {\sqrt {4 a c - b^2} } } + C & : b^2 - 4 a c < 0 \\ \dfrac 1 {\sqrt c} \ln \size {2 c \paren {\frac 1 x} + b} + C & : b^2 - 4 a c = 0 \end{cases}\) substituting for $u$
\(\ds \) \(=\) \(\ds \begin{cases} \dfrac 1 {\sqrt c} \ln \size {2 \sqrt c \sqrt {\dfrac {a x^2 + b x + c} {x^2} } + \dfrac {b x + 2 c} x} + C & : b^2 - 4 a c > 0 \\ \dfrac 1 {\sqrt c} \map {\sinh^{-1} } {\dfrac {b x + 2 c} {x \sqrt {4 a c - b^2} } } + C & : b^2 - 4 a c < 0 \\ \dfrac 1 {\sqrt c} \ln \size {\dfrac {2 c} x + b} + C & : b^2 - 4 a c = 0 \end{cases}\) simplifying
\(\ds \) \(=\) \(\ds \begin{cases} \dfrac {-1} {\sqrt c} \paren {-\ln \size {\dfrac {2 \sqrt c \sqrt {a x^2 + b x + c} + b x + 2 c} x} } + C & : b^2 - 4 a c > 0 \\ \dfrac {-1} {\sqrt c} \map {\sinh^{-1} } {\dfrac {b x + 2 c} {\paren {-x} \sqrt {4 a c - b^2} } } + C & : b^2 - 4 a c < 0 \\ \dfrac {-1} {\sqrt c} \paren {-\ln \size {\dfrac {2 c} x + b} } + C & : b^2 - 4 a c = 0 \end{cases}\)
\(\ds \) \(=\) \(\ds \begin{cases} \dfrac {-1} {\sqrt c} \dfrac {\size x} x \ln \size {\dfrac {2 \sqrt c \sqrt {a x^2 + b x + c} + b x + 2 c} x} + C & : b^2 - 4 a c > 0 \\ \dfrac {-1} {\sqrt c} \map {\sinh^{-1} } {\dfrac {b x + 2 c} {\size x \sqrt {4 a c - b^2} } } + C & : b^2 - 4 a c < 0 \\ \dfrac {-1} {\sqrt c} \dfrac {\size x} x \ln \size {\dfrac {2 c} x + b} + C & : b^2 - 4 a c = 0 \end{cases}\)

$\Box$


Finally we take $c < 0$.

Thus:

\(\ds \int \frac {\d u} {\sqrt {a + b u + c u^2} }\) \(=\) \(\ds \dfrac {-1} {\sqrt {-c} } \map \arcsin {\dfrac {2 c u + b} {\sqrt {\size {b^2 - 4 a c} } } }\) Primitive of $\dfrac 1 {\sqrt {a x^2 + b x + c} }$
\(\ds \leadsto \ \ \) \(\ds \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }\) \(=\) \(\ds \dfrac {-1} {\sqrt {-c} } \map \arcsin {\dfrac {2 c \paren {\frac 1 x} + b} {\sqrt {\size {b^2 - 4 a c} } } }\) substituting for $u$
\(\ds \) \(=\) \(\ds \dfrac 1 {\sqrt {-c} } \map \arcsin {\dfrac {b x + 2 c} {\paren {-x} \sqrt {\size {b^2 - 4 a c} } } }\) simplifying
\(\ds \) \(=\) \(\ds \dfrac 1 {\sqrt {-c} } \map \arcsin {\dfrac {b x + 2 c} {\size x \sqrt {\size {b^2 - 4 a c} } } }\) simplifying

$\blacksquare$


Proof 2 (incomplete)

\(\ds x \sqrt {a x^2 + b x + c}\) \(=\) \(\ds \frac x {\paren {a x^2 + b x + c}^{-\frac 1 2} }\)
\(\ds \) \(=\) \(\ds \frac{x \paren {2 \sqrt c \sqrt {a x^2 + b x + c} + b x + 2 c} } {\paren {a x^2 + b x + c}^{-\frac 1 2} \paren {2 \sqrt c \sqrt {a x^2 + b x + c} + b x + 2 c} }\)
\(\ds \) \(=\) \(\ds \frac{x \paren {2 \sqrt c \sqrt {a x^2 + b x + c} + b x + 2 c} } {2 \sqrt c + \paren {b x + 2 c} \paren {a x^2 + b x + c}^{-\frac 1 2} }\)
\(\ds \) \(=\) \(\ds \frac N D\)
where:
\(\ds N\) \(=\) \(\ds \frac {2 \sqrt c \sqrt {a x^2 + b x + c} + b x + 2 c} x\)
\(\ds D\) \(=\) \(\ds \frac {2 \sqrt c + \paren {b x + 2 c} \paren {a x^2 + b x + c}^{-\frac 1 2} } {x^2}\)

Then:

\(\ds \frac {\d N} {\d x}\) \(=\) \(\ds \frac {x \paren {\sqrt c \paren {2 a x + b} \paren {a x^2 + b x + c}^{-\frac 1 2} + b} - \paren {2 \sqrt c \sqrt {a x^2 + b x + c} + b x + 2 c} } {x^2}\)
\(\ds \) \(=\) \(\ds \paren {a x^2 + b x + c}^{-\frac 1 2} \frac {\paren {\sqrt c \paren {2 a x^2 + b x} + b x \sqrt {a x^2 + b x + c} } - \paren {2 \sqrt c \paren {a x^2 + b x + c} + \paren {b x + 2 c} \sqrt {a x^2 + b x + c} } } {x^2}\)
\(\ds \) \(=\) \(\ds \paren {a x^2 + b x + c}^{-\frac 1 2} \frac {\paren {\sqrt c \paren {2 a x^2 + b x - 2 a x^2 - 2 b x - 2 c} + b x \sqrt {a x^2 + b x + c} } - \paren {\paren {b x + 2 c} \sqrt {a x^2 + b x + c} } } {x^2}\)
\(\ds \) \(=\) \(\ds \paren {a x^2 + b x + c}^{-\frac 1 2} \frac {\paren {\sqrt c \paren {-b x - 2 c} - \paren {\paren {2 c} \sqrt {a x^2 + b x + c} } } } {x^2}\)
\(\ds \) \(=\) \(\ds -\frac {2 c + \sqrt c \paren {b x + 2 c} \paren {a x^2 + b x + c}^{-\frac 1 2} } {x^2}\)
\(\ds \) \(=\) \(\ds \sqrt c \times D\)


Now provided $N$ is real and non zero:

\(\ds \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }\) \(=\) \(\ds \int \frac D N \rd x\)
\(\ds \) \(=\) \(\ds \frac {-1} {\sqrt c} \int \frac {\frac {\d N} {\d x} } N \rd x\)
\(\ds \) \(=\) \(\ds \frac {-1} {\sqrt c} \ln \size N\) Primitive of Function under its Derivative
\(\ds \) \(=\) \(\ds \frac {-1} {\sqrt c} \ln \size {\frac {\paren {2 \sqrt c \sqrt {a x^2 + b x + c} + b x + 2 c} } x}\)


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