nyuuzyou/wikis huggingface dataset:
Primitive of Reciprocal of x cubed by x squared minus a squared/Partial Fraction Expansion

== Lemma for Primitive of Reciprocal of $x^3 \paren {x^2 - a^2}$ ==
:$\dfrac 1 {x^3 \paren {x^2 - a^2} } \equiv \dfrac {-1} {a^2 x^3} - \dfrac 1 {a^4 x} + \dfrac x {a^4 \paren {x^2 - a^2} }$
== Proof ==
Setting $x = 0$ in $(1)$:
Equating coefficients of $x$ in $(1)$:
Equating coefficients of $x^2$ in $(1)$:
Equating coefficients of $x^3$ in $(1)$:
Equating coefficients of $x^4$ in $(1)$:
Summarising:
Hence the result.
Category:Primitive of Reciprocal of x cubed by x squared minus a squared
Category:Primitives involving x squared minus a squared/Lemmata
Primitive of Reciprocal of x cubed by x squared minus a squared squared

== Theorem ==
:$\ds \int \frac {\d x} {x^3 \paren {x^2 - a^2}^2} = \frac {-1} {2 a^4 x^2} - \frac 1 {2 a^4 \paren {x^2 - a^2} } + \frac 1 {a^6} \map \ln {\frac {x^2} {x^2 - a^2} } + C$
for $x^2 > a^2$.
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {x^3 \paren {a^2 - x^2}^2}$
== Sources ==
* : $\S 14$: Integrals involving $x^2 - a^2$, $x^2 > a^2$: $14.157$
* : $\S 17$: Tables of Special Indefinite Integrals: $(7)$ Integrals Involving $x^2 - a^2$, $x^2 > a^2$: $17.7.14.$
Category:Primitive of Reciprocal of x cubed by x squared minus a squared squared
Category:Primitives involving x squared minus a squared
Primitive of Reciprocal of x cubed by x squared minus a squared squared/Partial Fraction Expansion

== Lemma for Primitive of Reciprocal of $x^3 \paren {x^2 - a^2}^2$ ==
:$\dfrac 1 {x^3 \paren {x^2 - a^2}^2} \equiv \dfrac 1 {a^4 x^3} + \dfrac 2 {a^6 x} - \dfrac {2 x} {a^6 \paren {x^2 - a^2} } + \dfrac x {a^4 \paren {x^2 - a^2}^2}$
== Proof ==
Setting $x = 0$ in $(1)$:
Setting $x = a$ in $(1)$:
Setting $x = -a$ in $(1)$:
Equating coefficients of $x$ in $(1)$:
Equating coefficients of $x^2$ in $(1)$:
Equating coefficients of $x^6$ in $(1)$:
Equating coefficients of $x^3$ in $(1)$:
Summarising:
Thus:
Hence the result.
Category:Primitive of Reciprocal of x cubed by x squared minus a squared squared
Category:Primitives involving x squared minus a squared/Lemmata
Primitive of Reciprocal of x cubed by x squared plus a squared

== Theorem ==
:$\ds \int \frac {\d x} {x^3 \paren {x^2 + a^2} } = -\frac 1 {2 a^2 x^2} - \frac 1 {2 a^4} \map \ln {\frac {x^2 + a^2} {x^2} } + C$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $x^2 + a^2$: $14.131$
* : $\S 17$: Tables of Special Indefinite Integrals: $(6)$ Integrals Involving $x^2 + a^2$: $17.6.7.$
Category:Primitive of Reciprocal of x cubed by x squared plus a squared
Category:Primitives involving x squared plus a squared
Primitive of Reciprocal of x cubed by x squared plus a squared/Partial Fraction Expansion

== Lemma for Primitive of $\dfrac 1 {x^3 \paren {x^2 + a^2} }$ ==
:$\dfrac 1 {x^3 \paren {x^2 + a^2} } \equiv \dfrac 1 {a^2 x^3} - \dfrac 1 {a^4 x} + \dfrac x {a^4 \paren {x^2 + a^2} }$
== Proof ==
Setting $x = 0$ in $(1)$:
Equating coefficients of $x$ in $(1)$:
Equating coefficients of $x^2$ in $(1)$:
Equating coefficients of $x^3$ in $(1)$:
Equating coefficients of $x^4$ in $(1)$:
Summarising:
Hence the result.
Category:Primitive of Reciprocal of x cubed by x squared plus a squared
Category:Primitives involving x squared plus a squared/Lemmata
Primitive of Reciprocal of x cubed by x squared plus a squared squared

== Theorem ==
:$\ds \int \frac {\d x} {x^3 \paren {x^2 + a^2}^2} = -\frac 1 {2 a^4 x^2} - \frac 1 {2 a^4 \paren {x^2 + a^2} } - \frac 1 {a^6} \map \ln {\frac {x^2} {x^2 + a^2} } + C$
== Proof ==
Let:
== Sources ==
* : $\S 14$: Integrals involving $x^2 + a^2$: $14.138$
* : $\S 17$: Tables of Special Indefinite Integrals: $(6)$ Integrals Involving $x^2 + a^2$: $17.6.14.$
Category:Primitive of Reciprocal of x cubed by x squared plus a squared squared
Category:Primitives involving x squared plus a squared
Primitive of Reciprocal of x cubed by x squared plus a squared squared/Partial Fraction Expansion

== Lemma for Primitive of Reciprocal of $x^3 \paren {x^2 + a^2}^2$ ==
:$\dfrac 1 {x^3 \paren {x^2 + a^2}^2} \equiv -\dfrac 2 {a^6 x} + \dfrac 1 {a^4 x^3} + \dfrac {2 x} {a^6 \paren {x^2 + a^2} } + \dfrac x {a^4 \paren {x^2 + a^2}^2}$
== Proof ==
Setting $x = 0$ in $(1)$:
Equating coefficients of $x^2$ in $(1)$:
Equating coefficients of $x^6$ in $(1)$:
Equating coefficients of $x$ in $(1)$:
Equating coefficients of $x^5$ in $(1)$:
Equating coefficients of $x^3$ in $(1)$:
Equating coefficients of $x^4$ in $(1)$:
Summarising:
Hence the result.
Category:Primitive of Reciprocal of x cubed by x squared plus a squared squared
Category:Primitives involving x squared plus a squared/Lemmata
Primitive of Reciprocal of x cubed plus a cubed

== Theorem ==
:$\ds \int \frac {\d x} {x^3 + a^3} = \frac 1 {6 a^2} \ln \size {\frac {\paren {x + a}^2} {x^2 - a x + a^2} } + \frac 1 {a^2 \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3}$ 
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $x^3 + a^3$: $14.299$
* : $\S 17$: Tables of Special Indefinite Integrals: $(14)$ Integrals Involving $x^3 + a^3$: $17.14.1.$
Category:Primitive of Reciprocal of x cubed plus a cubed
Category:Primitives involving x cubed plus a cubed
Primitive of Reciprocal of x cubed plus a cubed/Lemma

== Lemma for Primitive of Reciprocal of $x^3 + a^3$ ==
:$\ds \int \frac {\d x} {x^2 - a x + a^2} = \frac 2 {a \sqrt 3} \map \arctan {\frac {2 x - a} {a \sqrt 3} }$
== Proof ==
The discriminant of $x^2 - a x + a^2$ is:
Thus:
Category:Primitive of Reciprocal of x cubed plus a cubed
Category:Primitives involving x cubed plus a cubed/Lemmata
Primitive of Reciprocal of x cubed plus a cubed/Partial Fraction Expansion

== Lemma for Primitive of Reciprocal of $x^3 + a^3$ ==
:$\dfrac 1 {x^3 + a^3} = \dfrac 1 {3 a^2 \paren {x + a} } - \dfrac {x - 2 a} {3 a^2 \paren {x^2 - a x + a^2} }$
== Proof ==
Equating coefficients of $x^2$ in $(1)$:
Equating coefficients of $x$ in $(1)$:
Setting $x = 0$ in $(1)$:
Summarising:
Hence the result.
Category:Primitive of Reciprocal of x cubed plus a cubed
Category:Primitives involving x cubed plus a cubed/Lemmata
Primitive of Reciprocal of x cubed plus a cubed squared

== Theorem ==
:$\ds \int \frac {\d x} {\paren {x^3 + a^3}^2} = \frac x {3 a^3 \paren {x^3 + a^3} } + \frac 1 {9 a^5} \map \ln {\frac {\paren {x + a}^2} {x^2 - a x + a^2} } + \frac 2 {3 a^5 \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3}$ 
== Proof ==
Then from Primitive of $\dfrac 1 {\paren {a x + b}^m \paren {p x + q}^n}$:
:$\ds \int \frac {\d x} {\paren {a x + b}^m \paren {p x + q}^n} = \frac {-1} {\paren {n - 1} \paren {b p - a q} } \paren {\frac 1 {\paren {a x + b}^{m - 1} \paren {p x + q}^{n - 1} } + a \paren {m + n - 2} \int \frac {\d x} {\paren {a x + b}^m \paren {p x + q}^{n - 1} } }$
Here we have $a = 1, b = 0, m = \dfrac 2 3, p = 1, q = a^3, n = 2$.
So:
== Sources ==
* : $\S 14$: Integrals involving $x^3 + a^3$: $14.304$
Category:Primitives involving x cubed plus a cubed
Primitive of Reciprocal of x fourth minus a fourth

== Theorem ==
:$\ds \int \frac {\d x} {x^4 - a^4} = \frac 1 {4 a^3} \ln \size {\frac {x - a} {x + a} } - \frac 1 {2 a^3} \arctan \frac x a$ 
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $x^4 \pm a^4$: $14.318$
Category:Primitives involving x to the fourth plus or minus a to the fourth
Primitive of Reciprocal of x fourth plus a fourth

== Theorem ==
:$\ds \int \frac {\d x} {x^4 + a^4} = \frac 1 {4 a^3 \sqrt 2} \map \ln {\frac {x^2 + a x \sqrt 2 + a^2} {x^2 - a x \sqrt 2 + a^2} } - \frac 1 {2 a^3 \sqrt 2} \paren {\map \arctan {1 - \frac {x \sqrt 2} a} - \map \arctan {1 + \frac {x \sqrt 2} a} }$
== Proof ==
Then:
Similarly:
Thus:
== Sources ==
* : $\S 14$: Integrals involving $x^4 \pm a^4$: $14.311$
* : $\S 17$: Tables of Special Indefinite Integrals: $(15)$ Integrals Involving $x^4 \pm a^4$: $17.15.1.$
Category:Primitives involving x to the fourth plus or minus a to the fourth
Primitive of Reciprocal of x fourth plus a fourth/Lemma 1

== Lemma for Primitive of Reciprocal of $x^4 + a^4$ ==
:$\ds \int \frac {\d x} {x^2 + a x \sqrt 2 + a^2} = \frac {\sqrt 2} a \map \arctan {1 + \frac {x \sqrt 2} a}$
== Proof ==
The discriminant of $x^2 + a x \sqrt 2 + a^2$ is:
Thus:
Category:Primitives involving x to the fourth plus or minus a to the fourth/Lemmata
Primitive of Reciprocal of x fourth plus a fourth/Lemma 2

== Lemma for Primitive of Reciprocal of $x^4 + a^4$ ==
:$\ds \int \frac {\d x} {x^2 - a x \sqrt 2 + a^2} = \frac {-\sqrt 2} a \map \arctan {1 - \frac {x \sqrt 2} a}$
== Proof ==
The discriminant of $x^2 - a x \sqrt 2 + a^2$ is:
Thus:
Category:Primitives involving x to the fourth plus or minus a to the fourth/Lemmata
Primitive of Reciprocal of x fourth plus a fourth/Partial Fraction Expansion

== Lemma for Primitive of Reciprocal of $x^4 + a^4$ ==
:$\dfrac 1 {x^4 + a^4} = \dfrac {x + a \sqrt 2} {2 a^3 \sqrt 2 \paren {x^2 + a x \sqrt 2 + a^2} } - \dfrac {x - a \sqrt 2} {2 a^3 \sqrt 2 \paren {x^2 - a x \sqrt 2 + a^2} }$
== Proof ==
Equating coefficients of $x^3$ in $(1)$:
Equating coefficients of $x^2$ in $(1)$:
Equating coefficients of $x$ in $(1)$:
Setting $x = 0$ in $(1)$:
Summarising:
Thus:
Category:Primitives involving x to the fourth plus or minus a to the fourth/Lemmata
Primitive of Reciprocal of x squared by Cube of Root of a x squared plus b x plus c

== Theorem ==
Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int \frac {\d x} {x^2 \paren {\sqrt {a x^2 + b x + c} }^3} = -\frac {a x^2 + 2 b x + c} {c^2 x \sqrt {a x^2 + b x + c} } + \frac {b^2 - 2 a c} {2 c^2} \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3} - \frac {3 b} {2 c^2} \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }$ 
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a x^2 + b x + c}$: $14.294$
Category:Primitives involving Root of a x squared plus b x plus c
Primitive of Reciprocal of x squared by Root of a squared minus x squared

== Theorem ==
:$\ds \int \frac {\d x} {x^2 \sqrt {a^2 - x^2} } = \frac {-\sqrt {a^2 - x^2} } {a^2 x} + C$
== Proof ==
Let:
Using Primitive of $ \dfrac 1 {x^m \sqrt {a x + b} }$:
:$\ds \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} }$
Setting:
== Also see ==
* Primitive of $\dfrac 1 {x^2 \sqrt {x^2 + a^2} }$
* Primitive of $\dfrac 1 {x^2 \sqrt {x^2 - a^2} }$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a^2 - x^2}$: $14.242$
* : Front endpapers: A Brief Table of Integrals: $35$.
Category:Primitives involving Root of a squared minus x squared
Primitive of Reciprocal of x squared by Root of a squared minus x squared cubed

== Theorem ==
:$\ds \int \frac {\d x} {x^2 \paren {\sqrt {a^2 - x^2} }^3} = \frac {-\sqrt {a^2 - x^2} } {a^4 x} + \frac x {a^4 \sqrt {a^2 - x^2} } + C$
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {x^2 \paren {\sqrt {x^2 + a^2} }^3}$
* Primitive of $\dfrac 1 {x^2 \paren {\sqrt {x^2 - a^2} }^3}$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a^2 - x^2}$: $14.256$
Category:Primitives involving Root of a squared minus x squared
Primitive of Reciprocal of x squared by Root of a x + b

== Theorem ==
:$\ds \int \frac {\d x} {x^2 \sqrt {a x + b} } = -\frac {\sqrt {a x + b} } {b x} - \frac a {2 b} \int \frac {\d x} {x \sqrt {a x + b} }$
== Proof ==
From Primitive of Power of $x$ by Power of $a x + b$: Increment of Power of $x$:
:$\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^{m + 1} \paren {a x + b}^{n + 1} } {\paren {m + 1} b} - \frac {\paren {m + n + 2} a} {\paren {m + 1} b} \int x^{m + 1} \paren {a x + b}^n \rd x$
Setting $m = -2$ and $n = -\dfrac 1 2$:
== Also see ==
* Primitive of Reciprocal of $x$ by Root of $a x + b$ for $\ds \int \dfrac {\d x} {x \sqrt {a x + b} }$.
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a x + b}$: $14.88$
* : Front endpapers: A Brief Table of Integrals: $15$.
* : $\S 17$: Tables of Special Indefinite Integrals: $(2)$ Integrals Involving $\sqrt {a x + b}$: $17.2.5.$
Category:Primitives involving Root of a x + b
Category:Primitives involving Reciprocals
Primitive of Reciprocal of x squared by Root of a x squared plus b x plus c

== 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 ==
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a x^2 + bx + c}$: $14.284$
Category:Primitives involving Root of a x squared plus b x plus c
Primitive of Reciprocal of x squared by Root of x squared minus a squared

== Theorem ==
:$\ds \int \frac {\d x} {x^2 \sqrt {x^2 - a^2} } = \frac {\sqrt {x^2 - a^2} } {a^2 x} + C$
for $\size x > a$.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Primitive of $\dfrac 1 {x^2 \sqrt {x^2 + a^2} }$
* Primitive of $\dfrac 1 {x^2 \sqrt {a^2 - x^2} }$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {x^2 - a^2}$: $14.214$
* : Front endpapers: A Brief Table of Integrals: $46$.
Category:Primitive of Reciprocal of x squared by Root of x squared minus a squared
Category:Primitives involving Root of x squared minus a squared
Primitive of Reciprocal of x squared by Root of x squared minus a squared/Proof 1

== Theorem ==
== Proof ==
Let:
Then:
Hence:
== Also see ==
* Primitive of $\dfrac 1 {x^2 \sqrt {x^2 + a^2} }$
* Primitive of $\dfrac 1 {x^2 \sqrt {a^2 - x^2} }$
Category:Primitive of Reciprocal of x squared by Root of x squared minus a squared
Primitive of Reciprocal of x squared by Root of x squared minus a squared/Proof 2

== Theorem ==
== Proof ==
Let:
Using Primitive of $\dfrac 1 {u^m \sqrt {a u + b} }$:
:$\ds \int \frac {\d u} {u^m \sqrt {a u + b} } = -\frac {\sqrt {a u + b} } {\paren {m - 1} b u^{m - 1} } - \frac {\paren {2 m - 3} a} {\paren {2 m - 2} b} \int \frac {\d u} {u^{m - 1} \sqrt {a u + b} }$
Setting:
== Also see ==
* Primitive of $\dfrac 1 {x^2 \sqrt {x^2 + a^2} }$
* Primitive of $\dfrac 1 {x^2 \sqrt {a^2 - x^2} }$
Category:Primitive of Reciprocal of x squared by Root of x squared minus a squared
Primitive of Reciprocal of x squared by Root of x squared minus a squared cubed

== Theorem ==
:$\ds \int \frac {\d x} {x^2 \paren {\sqrt {x^2 - a^2} }^3} = \frac {-\sqrt {x^2 - a^2} } {a^4 x} - \frac x {a^4 \sqrt {x^2 - a^2} } + C$
for $\size x > a$.
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {x^2 \paren {\sqrt {x^2 + a^2} }^3}$
* Primitive of $\dfrac 1 {x^2 \paren {\sqrt {a^2 - x^2} }^3}$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {x^2 - a^2}$: $14.228$
Category:Primitives involving Root of x squared minus a squared
Primitive of Reciprocal of x squared by Root of x squared plus a squared

== Theorem ==
:$\ds \int \frac {\d x} {x^2 \sqrt {x^2 + a^2} } = -\frac {\sqrt {x^2 + a^2} } {a^2 x} + C$
== Proof ==
Let:
Using Primitive of $ \dfrac 1 {x^m \sqrt {a x + b} }$:
:$\ds \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} }$
Setting:
== Also see ==
* Primitive of $\dfrac 1 {x^2 \sqrt {x^2 - a^2} }$
* Primitive of $\dfrac 1 {x^2 \sqrt {a^2 - x^2} }$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {x^2 + a^2}$: $14.187$
* : Front endpapers: A Brief Table of Integrals: $27$.
Category:Primitives involving Root of x squared plus a squared
Primitive of Reciprocal of x squared by Root of x squared plus a squared cubed

== Theorem ==
:$\ds \int \frac {\d x} {x^2 \paren {\sqrt {x^2 + a^2} }^3} = \frac {-\sqrt {x^2 + a^2} } {a^4 x} - \frac x {a^4 \sqrt {x^2 + a^2} } + C$
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {x^2 \paren {\sqrt {x^2 - a^2} }^3}$
* Primitive of $\dfrac 1 {x^2 \paren {\sqrt {a^2 - x^2} }^3}$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {x^2 + a^2}$: $14.201$
Category:Primitives involving Root of x squared plus a squared
Primitive of Reciprocal of x squared by a squared minus x squared

== Theorem ==
:$\ds \int \frac {\d x} {x^2 \paren {a^2 - x^2} } = \frac {-1} {a^2 x} + \frac 1 {2 a^3} \map \ln {\frac {a + x} {a - x} } + C$
for $x^2 < a^2$.
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {x^2 \paren {x^2 - a^2} }$
== Sources ==
* : $\S 14$: Integrals involving $a^2 - x^2$, $x^2 < a^2$: $14.168$
Category:Primitives involving a squared minus x squared
Primitive of Reciprocal of x squared by a squared minus x squared/Partial Fraction Expansion

== Lemma for Primitive of Reciprocal of $x^2 \paren {a^2 - x^2}$ ==
:$\dfrac 1 {x^2 \paren {a^2 - x^2} } \equiv \dfrac 1 {a^2 \paren {a^2 - x^2} } + \dfrac 1 {a^2 x^2}$
== Proof ==
Setting $x = 0$ in $(1)$:
Equating coefficients of $x$ in $(1)$:
Equating coefficients of $x^3$ in $(1)$:
Equating coefficients of $x^2$ in $(1)$:
Summarising:
Hence the result.
Category:Primitives involving a squared minus x squared/Lemmata
Primitive of Reciprocal of x squared by a squared minus x squared squared

== Theorem ==
:$\ds \int \frac {\d x} {x^2 \paren {a^2 - x^2}^2} = \frac {-1} {a^4 x} + \frac x {2 a^4 \paren {a^2 - x^2} } + \frac 3 {4 a^5} \map \ln {\frac {a + x} {a - x} } + C$
for $x^2 < a^2$.
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {x^2 \paren {a^2 - x^2}^2}$
== Sources ==
* : $\S 14$: Integrals involving $a^2 - x^2$, $x^2 < a^2$: $14.175$
Category:Primitives involving a squared minus x squared
Primitive of Reciprocal of x squared by a squared minus x squared squared/Partial Fraction Expansion

== Lemma for Primitive of Reciprocal of $x^2 \paren {a^2 - x^2}^2$ ==
:$\dfrac 1 {x^2 \paren {a^2 - x^2}^2} \equiv \dfrac 1 {a^4 x^2} + \dfrac 3 {4 a^5 \paren {a + x} } - \dfrac 3 {4 a^5 \paren {a - x} } + \dfrac 1 {4 a^4 \paren {a + x}^2} + \dfrac 1 {4 a^4 \paren {a - x}^2}$
== Proof ==
Setting $x = a$ in $(1)$:
Setting $x = -a$ in $(1)$:
Setting $x = 0$ in $(1)$:
Equating coefficients of $x$ in $(1)$:
Equating coefficients of $x^5$ in $(1)$:
Equating coefficients of $x^2$ in $(1)$:
Summarising:
Hence the result.
Category:Primitives involving a squared minus x squared/Lemmata
Primitive of Reciprocal of x squared by a x + b

== Theorem ==
:$\ds \int \frac {\d x} {x^2 \paren {a x + b} } = -\frac 1 {b x} + \frac a {b^2} \ln \size {\frac {a x + b} x} + C$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 14$: Integrals involving $a x + b$: $14.64$
* : $\S 17$: Tables of Special Indefinite Integrals: $(1)$ Integrals Involving $a x + b$: $17.1.5.$
Category:Primitive of Reciprocal of x squared by a x + b‎
Category:Primitives involving a x + b
Category:Primitives involving Reciprocals
Primitive of Reciprocal of x squared by a x + b/Partial Fraction Expansion

== Lemma for Primitive of $\dfrac 1 {x^2 \paren {a x + b} }$ ==
:$\dfrac 1 {x^2 \paren {a x + b} } \equiv -\dfrac a {b^2 x} + \dfrac 1 {b x^2} + \dfrac {a^2} {b^2 \paren {a x + b} }$
== Proof ==
Setting $a x + b = 0$ in $(1)$:
Equating constants in $(1)$:
Equating $2$nd powers of $x$ in $(1)$:
Summarising:
Hence the result.
Category:Primitives involving a x + b/Lemmata
Primitive of Reciprocal of x squared by a x + b cubed

== Theorem ==
:$\ds \int \frac {\d x} {x^2 \paren {a x + b}^3} = \frac {-a} {2 b^2 \paren {a x + b}^2} - \frac {2 a} {b^3 \paren {a x + b} } - \frac 1 {b^3 x} + \frac {3 a} {b^4} \ln \size {\frac {a x + b} x} + C$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $a x + b$: $14.78$
Category:Primitives involving a x + b‎
Category:Primitives involving Reciprocals
Primitive of Reciprocal of x squared by a x + b cubed/Partial Fraction Expansion

== Lemma for Primitive of $\dfrac 1 {x^2 \paren {a x + b}^3}$ ==
:$\dfrac 1 {x^2 \paren {a x + b}^3} \equiv \dfrac {-3 a} {b^4 x} + \dfrac 1 {b^3 x^2} + \dfrac {3 a^2} {b^4 \paren {a x + b} } + \dfrac {2 a^2} {b^3 \paren {a x + b}^2} + \dfrac {a^2} {b^2 \paren {a x + b}^3}$
== Proof ==
Setting $a x + b = 0$ in $(1)$:
Equating constants in $(1)$:
Equating $1$st powers of $x$ in $(1)$:
Equating $4$th powers of $x$ in $(1)$:
Equating $3$rd powers of $x$ in $(1)$:
Summarising:
Hence the result.
Category:Primitives involving a x + b/Lemmata
Primitive of Reciprocal of x squared by a x + b squared

== Theorem ==
:$\ds \int \frac {\d x} {x^2 \paren {a x + b}^2} = \frac {-a} {b^2 \paren {a x + b} } - \frac 1 {b^2 x} + \frac {2 a} {b^3} \ln \size {\frac {a x + b} x} + C$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $a x + b$: $14.71$
* : $\S 17$: Tables of Special Indefinite Integrals: $(1)$ Integrals Involving $a x + b$: $17.1.10.$
Category:Primitives involving a x + b‎
Category:Primitives involving Reciprocals
Primitive of Reciprocal of x squared by a x + b squared/Partial Fraction Expansion

== Lemma for Primitive of $\dfrac 1 {x^2 \paren {a x + b}^2}$ ==
:$\dfrac 1 {x^2 \paren {a x + b}^2} \equiv -\dfrac {2 a} {b^3 x} + \dfrac 1 {b^2 x^2} + \dfrac {2 a^2} {b^3 \paren {a x + b} } + \dfrac {a^2} {b^2 \paren {a x + b}^2}$
== Proof ==
Setting $a x + b = 0$ in $(1)$:
Equating constants in $(1)$:
Equating $1$st powers of $x$ in $(1)$:
Equating $3$rd powers of $x$:
Summarising:
Hence the result.
Category:Primitives involving a x + b/Lemmata
Primitive of Reciprocal of x squared by a x squared plus b x plus c

== Theorem ==
Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int \frac {\d x} {x^2 \paren {a x^2 + b x + c} } = \frac b {2 c^2} \ln \size {\frac {a x^2 + b x + c} {x^2} } - \frac 1 {c x} + \frac {b^2 - 2 a c} {2 c^2} \int \frac {\d x} {a x^2 + b x + c}$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $a x^2 + b x + c$: $14.270$
Category:Primitives involving a x squared plus b x plus c
Primitive of Reciprocal of x squared by a x squared plus b x plus c/Partial Fraction Expansion

== Lemma for Primitive of $\dfrac 1 {x^2 \paren {a x^2 + b x + c} }$ ==
:$\dfrac 1 {x^2 \paren {a x^2 + b x + c} } \equiv \dfrac {-b} {c^2 x} + \dfrac 1 {c x^2} + \dfrac {a b x + b^2 - a c} {c^2 \paren {a x^2 + b x + c} }$
== Proof ==
Setting $x = 0$ in $(1)$:
Equating coefficients of $x$ in $(1)$:
Equating coefficients of $x^3$ in $(1)$:
Equating coefficients of $x^2$ in $(1)$:
Summarising:
Hence the result.
Category:Primitives involving a x squared plus b x plus c/Lemmata
Primitive of Reciprocal of x squared by square of a x squared plus b x plus c

== Theorem ==
Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int \frac {\d x} {x^2 \paren {a x^2 + b x + c}^2} = \frac {-1} {c x \paren {a x^2 + b x + c} } - \frac {3 a} c \int \frac {\d x} {\paren {a x^2 + b x + c}^2} - \frac {2 b} c \int \frac {\d x} {x \paren {a x^2 + b x + c}^2}$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 14$: Integrals involving $a x^2 + bx + c$: $14.278$
Category:Primitives involving a x squared plus b x plus c
Category:Primitive of Reciprocal of x squared by square of a x squared plus b x plus c
Primitive of Reciprocal of x squared by square of a x squared plus b x plus c/Proof 2

== Theorem ==
== Proof ==
First:
Next, with a view to obtaining an expression in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
and let:
Then:
Thus:
Category:Primitive of Reciprocal of x squared by square of a x squared plus b x plus c
Primitive of Reciprocal of x squared by x cubed plus a cubed

== Theorem ==
:$\ds \int \frac {\d x} {x^2 \paren {x^3 + a^3} } = \frac {-1} {a^3 x} - \frac 1 {6 a^4} \map \ln {\frac {x^2 - a x + a^2} {\paren {x + a}^2} } - \frac 1 {a^4 \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3}$ 
== Proof ==
First a lemma:
=== Lemma ===
Then:
== Sources ==
* : $\S 14$: Integrals involving $x^3 + a^3$: $14.303$
Category:Primitives involving x cubed plus a cubed
Category:Primitive of Reciprocal of x squared by x cubed plus a cubed
Primitive of Reciprocal of x squared by x cubed plus a cubed/Lemma

== Theorem ==
:$\ds \int \frac {\d x} {x^2 \paren {x^3 + a^3} } = \frac {-1} {a^3 x} - \frac 1 {a^3} \int \frac {x \rd x} {\paren {x^3 + a^3} }$ 
== Proof ==
Category:Primitives involving x cubed plus a cubed/Lemmata
Category:Primitive of Reciprocal of x squared by x cubed plus a cubed
Primitive of Reciprocal of x squared by x cubed plus a cubed squared

== Theorem ==
:$\ds \int \frac {\d x} {x^2 \paren {x^3 + a^3}^2} = \frac {-1} {a^6 x} - \frac {x^2} {3 a^6 \paren {x^3 + a^3} } - \frac 4 {3 a^6} \int \frac {x \rd x} {x^3 + a^3}$ 
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $x^3 + a^3$: $14.308$
Category:Primitives involving x cubed plus a cubed