Primitives involving a x squared plus b x plus c

Theorem

This page gathers together the primitives of some expressions involving $a x^2 + b x + c$.

Primitive of Reciprocal of $a x^2 + b x + c$

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

\dfrac 2 {\sqrt {4 a c - b^2} } \map \arctan {\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } } + C & : b^2 - 4 a c < 0 \\ \dfrac 1 {\sqrt {b^2 - 4 a c} } \ln \size {\dfrac {2 a x + b - \sqrt {b^2 - 4 a c} } {2 a x + b + \sqrt {b^2 - 4 a c} } } + C & : b^2 - 4 a c > 0 \\ \dfrac {-2} {2 a x + b} + C & : b^2 = 4 a c \end {cases}$

Primitive of $x$ over $a x^2 + b x + c$

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

Then:

$\ds \int \frac {x \rd x} {a x^2 + b x + c} = \frac 1 {2 a} \ln \size {a x^2 + b x + c} - \frac b {2 a} \int \frac {\d x} {a x^2 + b x + c}$

Primitive of $x^2$ over $a x^2 + b x + c$

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

Then:

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

Primitive of $x^m$ over $a x^2 + b x + c$

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

Then:

$\ds \int \frac {x^m \rd x} {a x^2 + b x + c} = \frac {x^{m - 1} } {\paren {m - 1} a} - \frac b a \int \frac {x^{m - 1} \rd x} {a x^2 + b x + c} - \frac c a \int \frac {x^{m - 2} \rd x} {a x^2 + b x + c}$

Primitive of Reciprocal of $x \left({a x^2 + b x + c}\right)$

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

Then:

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

Primitive of Reciprocal of $x^2 \left({a x^2 + b x + c}\right)$

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

Then:

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

Primitive of Reciprocal of $x^n \left({a x^2 + b x + c}\right)$

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

Then:

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

Primitive of Reciprocal of $\left({a x^2 + b x + c}\right)^2$

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

Then:

$\ds \int \frac {\d x} {\paren {a x^2 + b x + c}^2} = \frac {2 a x + b} {\paren {4 a c - b^2} \paren {a x^2 + b x + c} } + \frac {2 a} {4 a c - b^2} \int \frac {\d x} {a x^2 + b x + c}$

Primitive of $x$ over $\left({a x^2 + b x + c}\right)^2$

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

Then:

$\ds \int \frac {x \rd x} {\paren {a x^2 + b x + c}^2} = \frac {-\paren {b x + 2 c} } {\paren {4 a c - b^2} \paren {a x^2 + b x + c} } - \frac b {4 a c - b^2} \int \frac {\d x} {a x^2 + b x + c}$

Primitive of $x^2$ over $\left({a x^2 + b x + c}\right)^2$

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

Then:

$\ds \int \frac {x^2 \rd x} {\paren {a x^2 + b x + c}^2} = \frac {\paren {b^2 - 2 a c} x + b c} {a \paren {4 a c - b^2} \paren {a x^2 + b x + c} } + \frac {2 c} {4 a c - b^2} \int \frac {\d x} {a x^2 + b x + c}$

Primitive of $x^m$ over $\left({a x^2 + b x + c}\right)^n$

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

Then:

\(\ds \int \frac {x^m \rd x} {\paren {a x^2 + b x + c}^n}\)

\(=\)

\(\ds \frac {x^{m - 1} } {\paren {2 n - m - 1} a \paren {a x^2 + b x + c}^{n - 1} }\)

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds \frac {\paren {m - 1} c} {\paren {2 n - m - 1} a} \int \frac {x^{m - 2} \rd x} {\paren {a x^2 + b x + c}^n}\)

\(\ds \)

\(\)

\(\, \ds - \, \)

\(\ds \frac {\paren {n - m} b} {\paren {2 n - m - 1} a} \int \frac {x^{m - 1} \rd x} {\paren {a x^2 + b x + c}^n}\)

Primitive of $x^{2 n - 1}$ over $\left({a x^2 + b x + c}\right)^n$

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

Then:

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

Primitive of Reciprocal of $x \left({a x^2 + b x + c}\right)^2$

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

Then:

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

Primitive of Reciprocal of $x^2 \left({a x^2 + b x + c}\right)^2$

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

Then:

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

Primitive of Reciprocal of $x^m \left({a x^2 + b x + c}\right)^n$

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

Then:

\(\ds \int \frac {\d x} {x^m \paren {a x^2 + b x + c}^n}\)

\(=\)

\(\ds \frac {-1} {\paren {m - 1} c x^{m - 1} \paren {a x^2 + b x + c}^{n - 1} }\)

\(\ds \)

\(\)

\(\, \ds - \, \)

\(\ds \frac {\paren {m + 2 n - 3} a} {\paren {m - 1} c} \int \frac {\d x} {x^{m - 2} \paren {a x^2 + b x + c}^n}\)

\(\ds \)

\(\)

\(\, \ds - \, \)

\(\ds \frac {\paren {m - n + 2} b} {\paren {m - 1} c} \int \frac {\d x} {x^{m - 1} \paren {a x^2 + b x + c}^n}\)

Also see

• Primitives involving $\sqrt{a x^2 + b x + c}$