hcju/proofwikips huggingface dataset:
Primitive of x over a x + b cubed
:$\displaystyle \int \frac {x \ \mathrm d x} {\left({a x + b}\right)^3} = \frac {-1} {a^2 \left({a x + b}\right)} + \frac b {2 a^2 \left({a x + b}\right)^2} + C$
Primitive of Reciprocal of x by a x + b cubed
:$\displaystyle \int \frac {\mathrm d x} {x \left({a x + b}\right)^3} = \frac {a^2 x^2} {2 b^3 \left({a x + b}\right)^2} - \frac {2 a x} {b^3 \left({a x + b}\right)} + \frac 1 {b^3} \ln \left\vert{\frac x {a x + b} }\right\vert + C$
Primitive of Reciprocal of x squared by a x + b cubed
:$\displaystyle \int \frac {\mathrm d x} {x^2 \left({a x + b}\right)^3} = \frac {-a} {2 b^2 \left({a x + b}\right)^2} - \frac {2 a} {b^3 \left({a x + b}\right)} - \frac 1 {b^3 x} + \frac {3 a} {b^4} \ln \left\vert{\frac {a x + b} x}\right\vert + C$
Primitive of Reciprocal of x cubed by a x + b cubed
:$\displaystyle \int \frac {\d x} {x^3 \paren {a x + b}^3} = \frac {a^4 x^2} {2 b^5 \paren {a x + b}^2} - \frac {4 a^3 x} {b^5 \paren {a x + b} } - \frac {\paren {a x + b}^2} {2 b^5 x^2} + \frac {4 a} {b^4 x} + \frac {6 a^2} {b^5} \ln \size {\frac x {a x + b} } + C$
Laplace Transform of Second Derivative
Let $f: \R \to \R$ or $\R \to \C$ be a continuous function on any interval of the form $0 \le t \le a$. Let $f$ be twice differentiable. Let $f'$ be continuous and $f''$ piecewise continuous with one-sided limits on said intervals. Let $f$ and $f'$ be of exponential order. Let $\laptrans f$ denote the Laplace transform of $f$. Then $\laptrans {f''}$ exists for $\map \Re s > a$, and: :$\laptrans {\map {f''} t} = s^2 \laptrans {\map f t} - s \, \map f 0 - \map {f'} 0$
Euler's Number is Transcendental
Euler's Number $e$ is transcendental.
Laplace Transform of Hyperbolic Cosine
Let $\cosh t$ be the hyperbolic cosine, where $t$ is real. Let $\laptrans f$ denote the Laplace transform of the real function $f$. Then: :$\laptrans {\cosh a t} = \dfrac s {s^2 - a^2}$ where $a \in \R_{>0}$ is constant, and $\map \Re s > a$.
Laplace Transform of Hyperbolic Sine
Let $\sinh t$ be the hyperbolic sine, where $t$ is real. Let $\laptrans f$ denote the Laplace transform of the real function $f$. Then: :$\laptrans {\sinh a t} = \dfrac a {s^2 - a^2}$ where $a \in \R_{>0}$ is constant, and $\map \Re s > a$.
Primitives of Rational Functions involving a x + b cubed
This page gathers together the primitives of some rational functions involving $\left({a x + b}\right)^3$.
Primitive of Power of a x + b
:$\displaystyle \int \paren {a x + b}^n \rd x = \frac {\paren {a x + b}^{n + 1} } {\paren {n + 1} a} + C$
Primitive of x by Power of a x + b
:$\displaystyle \int x \left({a x + b}\right)^n \ \mathrm d x = \frac {\left({a x + b}\right)^{n + 2} } {\left({n + 2}\right) a^2} - \frac {b \left({a x + b}\right)^{n + 1} } {\left({n + 1}\right) a^2} + C$
Primitive of x squared by Power of a x + b
:$\displaystyle \int x^2 \left({a x + b}\right)^n \ \mathrm d x = \frac {\left({a x + b}\right)^{n + 3} } {\left({n + 3}\right) a^3} - \frac {2 b \left({a x + b}\right)^{n + 2} } {\left({n + 2}\right) a^3} + \frac {b^2 \left({a x + b}\right)^{n + 1} } {\left({n + 1}\right) a^3} + C$
Reduction Formula for Primitive of Power of x by Power of a x + b/Decrement of Power of a x + b
:$\displaystyle \int x^m \left({a x + b}\right)^n \rd x = \frac {x^{m+1} \left({a x + b}\right)^n} {m + n + 1} + \frac {n b} {m + n + 1} \int x^m \left({a x + b}\right)^{n - 1} \rd x$
Primitive of x over a x + b cubed/Proof 1
:$\displaystyle \int \frac {x \ \mathrm d x} {\left({a x + b}\right)^3} = \frac {-1} {a^2 \left({a x + b}\right)} + \frac b {2 a^2 \left({a x + b}\right)^2} + C$
Primitive of x over a x + b cubed/Proof 2
:$\displaystyle \int \frac {x \ \mathrm d x} {\left({a x + b}\right)^3} = \frac {-1} {a^2 \left({a x + b}\right)} + \frac b {2 a^2 \left({a x + b}\right)^2} + C$
Reduction Formula for Primitive of Power of x by Power of a x + b
=== Decrement of Power of $a x + b$ === {{:Reduction Formula for Primitive of Power of x by Power of a x + b/Decrement of Power of a x + b}} === Decrement of Power of $x$ === {{:Reduction Formula for Primitive of Power of x by Power of a x + b/Decrement of Power of x}} === Increment of Power of $a x + b$ === {{:Reduction Formula for Primitive of Power of x by Power of a x + b/Increment of Power of a x + b}} === Increment of Power of $x$ === {{:Reduction Formula for Primitive of Power of x by Power of a x + b/Increment of Power of x}}
Distance from Subset to Element
Let $\struct {M, d}$ be a metric space. Let $S \subseteq M$ be a subset of $M$. Let $s \in S$. Then: :$\map d {s, S} = 0$ where $\map d {s, S}$ denotes the distance between $s$ and $S$.
Reduction Formula for Primitive of Power of x by Power of a x + b/Decrement of Power of x
:$\displaystyle \int x^m \left({a x + b}\right)^n \rd x = \frac {x^m \left({a x + b}\right)^{n + 1} } {\left({m + n + 1}\right) a} - \frac {m b} {\left({m + n + 1}\right) a} \int x^{m - 1} \left({a x + b}\right)^n \rd x$
Distance from Subset to Supremum
Let $S \subseteq \R$ be a subset of the real numbers. Suppose that the supremum $\sup S$ of $S$ exists. Then: :$\map d {\sup S, S} = 0$ where $\map d {\sup S, S}$ is the distance between $\sup S$ and $S$.
Distance between Element and Subset is Nonnegative
Let $\struct {M, d}$ be a metric space. Let $x \in M$ and $S \subseteq M$. Then: :$\map d {x, S} \ge 0$ where $\map d {x, S}$ is the distance between $x$ and $S$.
Distance from Subset to Infimum
Let $S \subseteq \R$ be a subset of the real numbers. Suppose that the infimum $\inf S$ of $S$ exists. Then: :$\map d {\inf S, S} = 0$ where $\map d {\inf S, S}$ is the distance between $\inf S$ and $S$.
Reduction Formula for Primitive of Power of x by Power of a x + b/Increment of Power of a x + b
:$\displaystyle \int x^m \paren {a x + b}^n \rd x = \frac {-x^{m + 1} \paren {a x + b}^{n + 1} } {\paren {n + 1} b} + \frac {m + n + 2} {\paren {n + 1} b} \int x^m \paren {a x + b}^{n + 1} \rd x$
Reduction Formula for Primitive of Power of x by Power of a x + b/Increment of Power of x
:$\displaystyle \int x^m \left({a x + b}\right)^n \rd x = \frac {x^{m+1} \left({a x + b}\right)^{n + 1} } {\left({m + 1}\right) b} - \frac {\left({m + n + 2}\right) a} {\left({m + 1}\right) b} \int x^{m + 1} \left({a x + b}\right)^n \rd x$
Primitive of Reciprocal of Root of a x + b
:$\displaystyle \int \frac {\d x} {\sqrt{a x + b} } = \frac {2 \sqrt {a x + b} } a + C$
Primitive of Reciprocal of x squared minus a squared/Inverse Hyperbolic Cotangent Form
:$\ds \int \frac {\d x} {x^2 - a^2} = -\frac 1 a \coth^{-1} {\frac x a} + C$ where $\size x > a$.
Primitive of Reciprocal of x by Root of a x + b
:$\displaystyle \int \frac {\d x} {x \sqrt{a x + b} } = \begin {cases} \dfrac 1 {\sqrt b} \map \ln {\dfrac {\sqrt {a x + b} - \sqrt b} {\sqrt {a x + b} + \sqrt b} } + C & : b > 0 \\ \dfrac 2 {\sqrt {-b} } \arctan \sqrt {\dfrac {a x + b} {-b} } + C & : b < 0 \end {cases}$
Primitive of Reciprocal of x squared by Root of a x + b
:$\displaystyle \int \frac {\d x} {x^2 \sqrt{a x + b} } = -\frac {\sqrt{a x + b} } {b x} - \frac a {2 b} \int \frac {\mathrm d x} {x \sqrt{a x + b} }$
Primitive of Root of a x + b
:$\displaystyle \int \sqrt{a x + b} \ \mathrm d x = \frac {2 \sqrt{\left({a x + b}\right)^3} } {3 a}$
Primitive of Reciprocal of a squared minus x squared/Inverse Hyperbolic Tangent Form
:$\ds \int \frac {\d x} {a^2 - x^2} = \frac 1 a \tanh^{-1} \frac x a + C$ where $\size x < a$.
Primitive of Reciprocal of Root of x squared plus a squared/Inverse Hyperbolic Sine Form
:$\ds \int \frac {\d x} {\sqrt {x^2 + a^2} } = \sinh^{-1} {\frac x a} + C$
Primitive of Reciprocal of Root of x squared minus a squared/Inverse Hyperbolic Cosine Form
:$\displaystyle \int \frac {\d x} {\sqrt {x^2 - a^2} } = \dfrac {\size x} x \cosh^{-1} {\size {\frac x a} } + C$ for $x > a$.
Primitive of Reciprocal of x by Root of x squared plus a squared/Inverse Hyperbolic Cosecant Form
:$\displaystyle \int \frac {\d x} {x \sqrt {x^2 + a^2} } = -\frac 1 a \csch^{-1} {\frac x a} + C$
Primitive of Reciprocal of x by Root of a squared minus x squared/Inverse Hyperbolic Secant Form
:$\displaystyle \int \frac {\d x} {x \sqrt {a^2 - x^2} } = -\frac 1 a \sech^{-1} {\frac x a} + C$
Primitive of x by Root of a x + b
:$\displaystyle \int x \sqrt{a x + b} \ \mathrm d x = \frac {2 \left({3 a x - 2 b}\right) } {15 a^2} \sqrt{\left({a x + b}\right)^3}$
Laplace Transform of Heaviside Step Function times Function
Let $\map f t: \R \to \R$ or $\R \to \C$ be a function of exponential order $a$ for some constant $a \in \R$. Let $f$ be piecewise continuous with one-sided limits on any closed interval of the form $\closedint 0 b$ where $b > 0$. Let $\map {u_c} t$ be the Heaviside step function. Let $\laptrans {\map f t} = \map F s$ denote the Laplace transform of $f$. Then: :$\laptrans {\map {u_c} t \, \map f {t - c} } = e^{-s c} \map F s$ for $\map \Re s > a$.
Primitive of x squared by Root of a x + b
:$\displaystyle \int x^2 \sqrt{a x + b} \ \mathrm d x = \frac {2 \left({15 a^2 x^2 - 12 a b x + 8 b^2}\right) } {105 a^3} \sqrt{\left({a x + b}\right)^3} + C$
Primitive of Root of a x + b over x
:$\displaystyle \int \frac {\sqrt {a x + b} } x \rd x = 2 \sqrt {a x + b} + b \int \frac {\d x} {x \sqrt{a x + b} }$
Primitive of Root of a x + b over x squared
:$\displaystyle \int \frac {\sqrt {a x + b} } {x^2} \rd x = -\frac {\sqrt {a x + b} } x + \frac a 2 \int \frac {\d x} {x \sqrt {a x + b} }$
Primitive of Power of x over Root of a x + b
:$\ds \int \frac {x^m} {\sqrt{a x + b} } \rd x = \frac {2 x^m \sqrt{a x + b} } {\paren {2 m + 1} a} - \frac {2 m b} {\paren {2 m + 1} a} \int \frac {x^{m - 1} } {\sqrt{a x + b} } \rd x$
Primitive of Reciprocal of Power of x by Root of a x + b
:$\displaystyle \int \frac {\d x} {x^m \sqrt {a x + b} } = -\frac {\sqrt {a x + b} } {\paren {m - 1} b x^{m - 1} } - \frac {\paren {2 m - 3} a} {\paren {2 m - 2} b} \int \frac {\d x} {x^{m - 1} \sqrt {a x + b} }$
Primitive of Root of a x + b over Power of x
=== Formulation 1 === {{:Primitive of Root of a x + b over Power of x/Formulation 1}} === Formulation 2 === {{:Primitive of Root of a x + b over Power of x/Formulation 2}}
Primitive of Power of Root of a x + b over x
:$\displaystyle \int \frac {\left({\sqrt{a x + b} }\right)^m} x \ \mathrm d x = \frac {2 \left({\sqrt{a x + b} }\right)^m } m + b \int \frac {\left({\sqrt{a x + b} }\right)^{m - 2} } x \ \mathrm d x$
Primitive of Reciprocal of x by Power of Root of a x + b
:$\displaystyle \int \frac {\mathrm d x} {x \left({\sqrt{a x + b} }\right)^m} = \frac 2 {\left({m - 2}\right) b \left({\sqrt{a x + b} }\right)^{m - 2} }  + \frac 1 b \int \frac {\mathrm d x} {x \left({\sqrt{a x + b} }\right)^{m - 2} }$
Primitive of Reciprocal of a x + b by p x + q
:$\displaystyle \int \frac {\mathrm d x} {\left({a x + b}\right) \left({p x + q}\right)} = \frac 1 {b p - a q} \ln \left\vert{\frac {p x + q} {a x + b} }\right\vert + C$
Primitive of x over a x + b by p x + q
:$\displaystyle \int \frac {x \ \mathrm d x} {\left({a x + b}\right) \left({p x + q}\right)} = \frac 1 {b p - a q} \left({\frac b a \ln \left\vert{a x + b}\right\vert - \frac q p \ln \left\vert{p x + q}\right\vert}\right) + C$
Primitive of Reciprocal of a x + b squared by p x + q
:$\displaystyle \int \frac {\mathrm d x} {\left({a x + b}\right)^2 \left({p x + q}\right)} = \frac 1 {b p - a q} \left({\frac 1 {a x + b} + \frac p {b p - a q} \ln \left\vert{\frac {p x + q} {a x + b} }\right\vert}\right) + C$
Primitive of x over a x + b squared by p x + q
:$\displaystyle \int \frac {x \rd x} {\paren {a x + b}^2 \paren {p x + q} } = \frac 1 {b p - a q} \paren {\frac q {b p - a q} \ln \size {\frac {a x + b} {p x + q} } - \frac b {a \paren {a x + b} } } + C$
Primitive of x squared over a x + b squared by p x + q
:$\displaystyle \int \frac {x^2 \ \mathrm d x} {\left({a x + b}\right)^2 \left({p x + q}\right)} = \frac {b^2} {\left({b p - a q}\right) a^2 \left({a x + b}\right)} + \frac 1 {\left({b p - a q}\right)^2} \left({\frac {q^2} p \ln \left\vert{p x + q}\right\vert + \frac {b \left({b p - 2 a q}\right)} {a^2} \ln \left\vert{a x + b}\right\vert}\right) + C$
Friedrichs' Inequality
Let $G \subset \R^n$ be bounded domain. Then for any $u \in \map {W^{2, 1}_0} G$: :$\norm u_{\map {L^2} G} \le \map {\operatorname {diam} } G \norm {\nabla u}_{\map {L^2} G}$ where: :$\map {\operatorname {diam} } G := \sup \limits_{x, y \mathop \in G} \size {x - y}$
Primitive of Reciprocal of Power of a x + b by Power of p x + q
:$\displaystyle \int \frac {\mathrm d x} {\left({a x + b}\right)^m \left({p x + q}\right)^n} = \frac {-1} {\left({n - 1}\right) \left({b p - a q}\right)} \left({\frac 1 {\left({a x + b}\right)^{m-1} \left({p x + q}\right)^{n-1} } + a \left({m + n - 2}\right) \int \frac {\mathrm d x} {\left({a x + b}\right)^m \left({p x + q}\right)^{n-1} } }\right)$
Reduction Formula for Primitive of Power of a x + b by Power of p x + q/Decrement of Power
:$\displaystyle \int \left({a x + b}\right)^m \left({p x + q}\right)^n \ \mathrm d x = \frac {\left({a x + b}\right)^{m+1} \left({p x + q}\right)^n} {\left({m + n + 1}\right) a} - \frac {n \left({b p - a q}\right)} {\left({m + n + 1}\right) a} \int \left({a x + b}\right)^m \left({p x + q}\right)^{n-1} \ \mathrm d x$
Reduction Formula for Primitive of Power of a x + b by Power of p x + q/Increment of Power
:$\displaystyle \int \left({a x + b}\right)^m \left({p x + q}\right)^n \ \mathrm d x = \frac 1 {\left({n + 1}\right) \left({b p - a q}\right)} \left({\left({a x + b}\right)^{m+1} \left({p x + q}\right)^{n+1} - a \left({m + n + 2}\right) \int \left({a x + b}\right)^m \left({p x + q}\right)^{n+1} \ \mathrm d x}\right)$
Primitive of a x + b over p x + q
:$\displaystyle \int \frac {a x + b} {p x + q} \ \mathrm d x  = \frac {a x} p + \frac {b p - a q} {p^2} \ln \left\vert{p x + q}\right\vert + C$
Reduction Formula for Primitive of Power of x by Power of a x + b/Decrement of Power of a x + b/Proof 1
:$\displaystyle \int x^m \left({a x + b}\right)^n \rd x = \frac {x^{m+1} \left({a x + b}\right)^n} {m + n + 1} + \frac {n b} {m + n + 1} \int x^m \left({a x + b}\right)^{n - 1} \rd x$
Primitive of Power of a x + b over Power of p x + q/Formulation 1
:$\displaystyle \int \frac {\left({a x + b}\right)^m} {\left({p x + q}\right)^n} \rd x = \frac {-1} {\left({n - 1}\right) \left({b p - a q}\right)} \left({\frac {\left({a x + b}\right)^{m + 1} } {\left({p x + q}\right)^{n - 1} } + \left({n - m - 2}\right) a \int \frac {\left({a x + b}\right)^m} { \left({p x + q}\right)^{n - 1} } \rd x}\right)$
Primitive of Power of a x + b over Power of p x + q
=== Formulation 1 === {{:Primitive of Power of a x + b over Power of p x + q/Formulation 1}} === Formulation 2 === {{:Primitive of Power of a x + b over Power of p x + q/Formulation 2}} === Formulation 3 === {{:Primitive of Power of a x + b over Power of p x + q/Formulation 3}}
Primitive of Power of a x + b over Power of p x + q/Formulation 2
:$\displaystyle \int \frac {\paren {a x + b}^m} {\paren {p x + q}^n} \rd x = \frac {-1} {\paren {n - m - 1} p} \paren {\frac {\paren {a x + b}^m} {\paren {p x + q}^{n - 1} } + m \paren {b p - a q} \int \frac {\paren {a x + b}^{m - 1} } {\paren {p x + q}^n} \rd x}$
Primitive of Power of a x + b over Power of p x + q/Formulation 3
:$\displaystyle \int \frac {\paren {a x + b}^m} {\paren {p x + q}^n} \rd x = \frac {-1} {\paren {n - 1} p} \paren {\frac {\paren {a x + b}^m} {\paren {p x + q}^{n - 1} } - m a \int \frac {\paren {a x + b}^{m - 1} } {\paren {p x + q}^{n - 1}} \rd x}$
Primitive of Reciprocal of p x + q by Root of a x + b
:$\displaystyle \int \frac {\d x} {\paren {p x + q} \sqrt {a x + b} } = \begin {cases} \dfrac 1 {\sqrt {b p - a q} \sqrt p} \ln \size {\dfrac {\sqrt {p \paren {a x + b} } - \sqrt {b p - a q} } {\sqrt {p \paren {a x + b} } + \sqrt {b p - a q} } } & : b p - a q > 0 \\ \dfrac 2 {\sqrt {a q - b p} \sqrt p} \arctan \sqrt {\dfrac {p \paren {a x + b} } {a q - b p} } & : b p - a q < 0 \\ \end {cases}$
Primitive of Root of a x + b over p x + q
:$\displaystyle \int \frac {\sqrt{a x + b} } {p x + q} \rd x = \begin{cases} \dfrac {2 \sqrt{a x + b} } p + \dfrac {\sqrt {b p - a q} } {p \sqrt p} \ln \size {\dfrac {\sqrt {p \paren {a x + b} } - \sqrt {b p - a q} } {\sqrt {p \paren {a x + b} } + \sqrt {b p - a q} } } & : b p - a q > 0 \\ \dfrac {2 \sqrt{a x + b} } p - \dfrac {\sqrt {a q - b p} } {p \sqrt p} \arctan \sqrt {\dfrac {p \paren {a x + b} } {a q - b p} } & : b p - a q < 0 \\ \end{cases}$
Primitive of Power of p x + q by Root of a x + b
:$\displaystyle \int \paren {p x + q}^n \sqrt {a x + b} \rd x = \frac {2 \paren {p x + q}^{n + 1} \sqrt {a x + b} } {\paren {2 n + 3} p} + \frac {b p - a q} {\paren {2 n + 3} p} \int \frac {\paren {p x + q}^n} {\sqrt{a x + b} } \rd x$
Primitive of Reciprocal of Power of p x + q by Root of a x + b
:$\displaystyle \int \frac {\mathrm d x} {\left({p x + q}\right)^n \sqrt{a x + b} } = \frac {\sqrt{a x + b} } {\left({n - 1}\right) \left({a q - b p}\right) \left({p x + q}\right)^{n-1} } + \frac {\left({2 n - 3}\right) a} {2 \left({n - 1}\right)  \left({a q - b p}\right)} \int \frac {\mathrm d x} {\left({p x + q}\right)^{n-1} \sqrt{a x + b} }$
Primitive of Power of p x + q over Root of a x + b
:$\displaystyle \int \frac {\left({p x + q}\right)^n} {\sqrt{a x + b} } \ \mathrm d x = \frac {2 \left({p x + q}\right)^n \sqrt{a x + b} } {\left({2 n + 1}\right) a} + \frac {2 n \left({a q - b p}\right)} {\left({2 n + 1}\right) a} \int \frac {\left({p x + q}\right)^{n-1} } {\sqrt{a x + b} } \ \mathrm d x$
Primitive of Reciprocal of Root of a x + b by Root of p x + q
:$\displaystyle \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } } = \begin {cases} \dfrac 2 {\sqrt {a p} } \map \ln {\sqrt {p \paren {a x + b} } + \sqrt {a \paren {p x + q} } } + C & : \dfrac {b p - a q} p > 0 \\ \dfrac 2 {\sqrt {a p} } \sinh^{-1} \sqrt {\dfrac {p \paren {a x + b} } {b p - a q} } + C & : \dfrac {b p - a q} p < 0 \\ \end {cases}$
Derivative of Laplace Transform
Let $f: \R \to \R$ or $\R \to \C$ be a continuous function, differentiable on any closed interval $\closedint 0 a$. Let $\laptrans f = F$ denote the Laplace transform of $f$. Then, everywhere that $\dfrac \d {\d s} \laptrans f$ exists: :$\dfrac \d {\d s} \laptrans {\map f t} = -\laptrans {t \, \map f t}$
Primitive of x over Root of a x + b by Root of p x + q
:$\displaystyle \int \frac {x \ \mathrm d x} {\sqrt{\left({a x + b}\right) \left({p x + q}\right)} } = \frac {\sqrt{\left({a x + b}\right) \left({p x + q}\right)} } {a p} - \frac {b p + a q} {2 a p} \int \frac {\mathrm d x} {\sqrt{\left({a x + b}\right) \left({p x + q}\right)} }$
Higher Order Derivatives of Laplace Transform
:$\dfrac {\d^n} {\d s^n} \laptrans {\map f t} = \paren {-1}^n \laptrans {t^n \, \map f t}$
Primitive of Root of a x + b by Root of p x + q
:$\displaystyle \int \sqrt {\paren {a x + b} \paren {p x + q} } \rd x = \frac {2 a p x + b p + a q} {4 a p} \sqrt {\paren {a x + b} \paren {p x + q} } - \frac {\paren {b p - a q}^2} {8 a p} \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }$
Primitive of Root of p x + q over Root of a x + b
:$\displaystyle \int \sqrt {\frac {p x + q} {a x + b} } \rd x = \frac {\sqrt {\paren {a x + b} \paren {p x + q} } } a + \frac {a q - b p} {2 a} \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }$
Primitive of Reciprocal of p x + q by Root of a x + b by Root of p x + q
:$\displaystyle \int \frac {\d x} {\paren {p x + q} \sqrt {\paren {a x + b} \paren {p x + q} } } = \frac {2 \sqrt{a x + b} } {\paren {a q - b p} \sqrt {p x + q} } + C$
Primitive of x over x squared plus a squared
:$\displaystyle \int \frac {x \rd x} {x^2 + a^2} = \frac 1 2 \map \ln {x^2 + a^2} + C$
Primitive of x squared over x squared plus a squared
:$\displaystyle \int \frac {x^2 \rd x} {x^2 + a^2} = x - a \arctan {\frac x a} + C$
Primitive of x cubed over x squared plus a squared
:$\displaystyle \int \frac {x^3 \rd x} {x^2 + a^2} = \frac {x^2} 2 - \frac {a^2} 2 \map \ln {x^2 + a^2} + C$
Primitive of Reciprocal of x by x squared plus a squared
:$\ds \int \frac {\rd x} {x \paren {x^2 + a^2} } = \frac 1 {2 a^2} \map \ln {\frac {x^2} {x^2 + a^2} } + C$
Primitive of Reciprocal of x squared by x squared plus a squared
:$\displaystyle \int \frac {\d x} {x^2 \paren {x^2 + a^2} } = -\frac 1 {a^2 x} - \frac 1 {a^3} \arctan \frac x a + C$
Primitive of Reciprocal of x cubed by x squared plus a squared
:$\displaystyle \int \frac {\mathrm d x} {x^3 \left({x^2 + a^2}\right)} = -\frac 1 {2 a^2 x^2} - \frac 1 {2 a^4} \ln \left({\frac {x^2 + a^2} {x^2} }\right) + C$
Primitive of Reciprocal of x squared plus a squared squared
:$\displaystyle \int \frac {\d x} {\paren {x^2 + a^2}^2} = \frac x {2 a^2 \paren {x^2 + a^2} } + \frac 1 {2 a^3} \arctan \frac x a + C$
Primitive of x over x squared plus a squared squared
:$\displaystyle \int \frac {x \ \mathrm d x} {\left({x^2 + a^2}\right)^2} = -\frac 1 {2 \left({x^2 + a^2}\right)} + C$
Primitive of x squared over x squared plus a squared squared
:$\displaystyle \int \frac {x^2 \ \mathrm d x} {\left({x^2 + a^2}\right)^2} = \frac {-x} {2 \left({x^2 + a^2}\right)} + \frac 1 {2 a} \arctan \frac x a + C$
Primitive of x cubed over x squared plus a squared squared
:$\displaystyle \int \frac {x^3 \ \mathrm d x} {\left({x^2 + a^2}\right)^2} = \frac {a^2} {2 \left({x^2 + a^2}\right)} + \frac 1 2 \ln \left({x^2 + a^2}\right) + C$
Primitive of Reciprocal of x by x squared plus a squared squared
:$\displaystyle \int \frac {\d x} {x \paren {x^2 + a^2}^2} = \frac 1 {2 a^2 \paren {x^2 + a^2} } + \frac 1 {2 a^4} \map \ln {\frac {x^2} {x^2 + a^2} } + C$
Primitive of Reciprocal of x squared by x squared plus a squared squared
:$\displaystyle \int \frac {\d x} {x^2 \paren {x^2 + a^2}^2} = -\frac 1 {a^4 x} - \frac x {2 a^4 \paren {x^2 + a^2} } - \frac 3 {2 a^5} \arctan \frac x a + C$
Primitive of Reciprocal of x cubed by x squared plus a squared squared
:$\displaystyle \int \frac {\d x} {x^3 \paren {x^2 + a^2}^2} = -\frac 1 {2 a^4 x^2} - \frac 1 {2 a^4 \paren {x^2 + a^2} } - \frac 1 {a^6} \map \ln {\frac {x^2} {x^2 + a^2} } + C$
Primitive of Reciprocal of Power of x squared plus a squared
:$\displaystyle \int \frac {\mathrm d x} {\left({x^2 + a^2}\right)^n} = \frac x {2 \left({n - 1}\right) a^2 \left({x^2 + a^2}\right)^{n - 1} } + \frac {2 n - 3} {\left({2 n - 2}\right) a^2} \int \frac {\mathrm d x} {\left({x^2 + a^2}\right)^{n - 1} }$
Primitive of Reciprocal of x by Power of x squared plus a squared
:$\displaystyle \int \frac {\mathrm d x} {x \left({x^2 + a^2}\right)^n} = \frac 1 {2 \left({n - 1}\right) a^2 \left({x^2 + a^2}\right)^{n - 1} } + \frac 1 {a^2} \int \frac {\mathrm d x} {x \left({x^2 + a^2}\right)^{n - 1} }$
Primitive of Reciprocal of Power of x by Power of x squared plus a squared
:$\displaystyle \int \frac {\d x} {x^m \paren {x^2 + a^2}^n} = \frac 1 {a^2} \int \frac {\d x} {x^m \paren {x^2 + a^2}^{n - 1} } - \frac 1 {a^2} \int \frac {\d x} {x^{m - 2} \paren {x^2 + a^2}^n}$
Primitive of x over x squared minus a squared
:$\displaystyle \int \frac {x \rd x} {x^2 - a^2} = \frac 1 2 \map \ln {x^2 - a^2} + C$ for $x^2 > a^2$.
Primitive of x cubed over x squared minus a squared
:$\displaystyle \int \frac {x^3 \ \mathrm d x} {x^2 - a^2} = \frac {x^2} 2 + \frac {a^2} 2 \ln \left({x^2 - a^2}\right) + C$ for $x^2 > a^2$.
Primitive of Reciprocal of x by x squared minus a squared
:$\displaystyle \int \frac {\d x} {x \paren {x^2 - a^2} } = \frac 1 {2 a^2} \, \map \ln {\frac {x^2 - a^2} {x^2} } + C$ for $x^2 > a^2$.
Primitive of Reciprocal of x squared by x squared minus a squared
:$\displaystyle \int \frac {\mathrm d x} {x^2 \left({x^2 - a^2}\right)} = \frac 1 {a^2 x} + \frac 1 {2 a^3} \ln \left({\frac {x - a} {x + a} }\right) + C$ for $x^2 > a^2$.
Primitive of Reciprocal of x cubed by x squared minus a squared
:$\displaystyle \int \frac {\mathrm d x} {x^3 \left({x^2 - a^2}\right)} = \frac 1 {2 a^2 x^2} + \frac 1 {2 a^4} \ln \left({\frac {x^2 - a^2} {x^2} }\right) + C$ for $x^2 > a^2$.
Primitive of Reciprocal of x squared minus a squared squared
:$\displaystyle \int \frac {\d x} {\left({x^2 - a^2}\right)^2} = \frac {-x} {2 a^2 \left({x^2 - a^2}\right)} + \frac 1 {4 a^3} \ln \left({\frac {x + a} {x - a} }\right) + C$ for $x^2 > a^2$.
Primitive of x over x squared minus a squared squared
:$\displaystyle \int \frac {x \ \mathrm d x} {\left({x^2 - a^2}\right)^2} = \frac {-1} {2 \left({x^2 - a^2}\right)} + C$ for $x^2 > a^2$.