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

== Lemma for Primitive of $\dfrac 1 {x \paren {a x + b}^3}$ ==
:$\dfrac 1 {x \paren {a x + b}^3} \equiv \dfrac 1 {b^3 x} - \dfrac a {b^3 \paren {a x + b} } - \dfrac a {b^2 \paren {a x + b}^2} - \dfrac a {b \paren {a x + b}^3}$
== Proof ==
Setting $a x + b = 0$ in $(1)$:
Equating constants in $(1)$:
Equating $3$rd powers of $x$ 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 by a x + b squared

== Theorem ==
:$\ds \int \frac {\d x} {x \paren {a x + b}^2} = \frac 1 {b \paren {a x + b} } + \frac 1 {b^2} \ln \size {\frac x {a x + b} } + C$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 14$: Integrals involving $a x + b$: $14.70$
* : $\S 17$: Tables of Special Indefinite Integrals: $(1)$ Integrals Involving $a x + b$: $17.1.9.$
Category:Primitive of Reciprocal of x by a x + b squared‎
Category:Primitives involving a x + b
Category:Primitives involving Reciprocals
Primitive of Reciprocal of x by a x + b squared/Partial Fraction Expansion

== Lemma for Primitive of $\dfrac 1 {x \paren {a x + b}^2}$ ==
:$\dfrac 1 {x \paren {a x + b}^2} \equiv \dfrac 1 {b^2 x} - \dfrac a {b^2 \paren {a x + b} } - \dfrac a {b \paren {a x + b}^2}$
== 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 by a x squared plus b x plus c

== Theorem ==
Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int \frac {\d x} {x \paren {a x^2 + b x + c} } = \frac 1 {2 c} \ln \size {\frac {x^2} {a x^2 + b x + c} } - \frac b {2 c} \int \frac {\d x} {a x^2 + b x + c}$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $a x^2 + bx + c$: $14.269$
Category:Primitives involving a x squared plus b x plus c
Primitive of Reciprocal of x by a x squared plus b x plus c/Partial Fraction Expansion

== Lemma for Primitive of $\dfrac 1 {x \paren {a x^2 + b x + c} }$ ==
:$\dfrac 1 {x \paren {a x^2 + b x + c} } \equiv \dfrac 1 {c x} - \dfrac {a x + b} {c \paren {a x^2 + b x + c} }$
== Proof ==
Setting $x = 0$ in $(1)$:
Equating coefficients of $x$ 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 by square of a x squared plus b x plus c

== Theorem ==
Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int \frac {\d x} {x \paren {a x^2 + b x + c}^2} = \frac 1 {2 c \paren {a x^2 + b x + c} } - \frac b {2 c} \int \frac {\d x} {\paren {a x^2 + b x + c}^2} + \frac 1 c \int \frac {\d x} {x \paren {a x^2 + b x + c} }$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $a x^2 + b x + c$: $14.277$
Category:Primitives involving a x squared plus b x plus c
Primitive of Reciprocal of x by square of a x squared plus b x plus c/Partial Fraction Expansion

== Lemma for Primitive of $\frac 1 {x \paren {a x^2 + b x + c}^2}$ ==
:$\dfrac 1 {x \paren {a x^2 + b x + c}^2} \equiv \dfrac 1 {c^2 x} - \dfrac {a x + b} {c^2 \paren {a x^2 + b x + c} } - \dfrac {a x + b} {c \paren {a x^2 + b x + c}^2}$
== Proof ==
Setting $x = 0$ in $(1)$:
Equating coefficients of $x^4$ in $(1)$:
Equating coefficients of $x^3$ in $(1)$:
Equating coefficients of $x$ 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 by x cubed plus a cubed

== Theorem ==
:$\ds \int \frac {\d x} {x \paren {x^3 + a^3} } = \frac 1 {3 a^3} \ln \size {\frac {x^3} {x^3 + a^3} } + C$ 
== Proof ==
From Primitive of $\dfrac 1 {x \paren {x^n + a^n} }$:
:$\ds \int \frac {\d x} {x \paren {x^n + a^n} } = \frac 1 {n a^n} \ln \size {\frac {x^n} {x^n + a^n} } + C$
Setting $n = 3$:
:$\ds \int \frac {\d x} {x \paren {x^3 + a^3} } = \frac 1 {3 a^3} \ln \size {\frac {x^3} {x^3 + a^3} } + C$
directly.
== Sources ==
* : $\S 14$: Integrals involving $x^3 + a^3$: $14.302$
Category:Primitives involving x cubed plus a cubed
Primitive of Reciprocal of x by x cubed plus a cubed squared

== Theorem ==
:$\ds \int \frac {\d x} {x \paren {x^3 + a^3}^2} = \frac 1 {3 a^3 \paren {x^3 + a^3} } + \frac 1 {3 a^6} \ln \size {\frac {x^3} {x^3 + a^3} }$ 
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $x^3 + a^3$: $14.307$
Category:Primitives involving x cubed plus a cubed
Primitive of Reciprocal of x by x fourth minus a fourth

== Theorem ==
:$\ds \int \frac {\d x} {x \paren {x^4 - a^4} } = \frac 1 {4 a^4} {\ln \size {\frac {x^4 - a^4} {x^4} } } + C$
== Proof ==
From Primitive of $\dfrac 1 {x \paren {x^n - a^n} }$:
:$\ds \int \frac {\d x} {x \paren {x^n - a^n} } = \frac 1 {n a^n} \ln \size {\frac {x^n - a^n} {x^n} } + C$
Setting $n = 4$:
:$\ds \int \frac {\d x} {x \paren {x^4 - a^4} } = \frac 1 {4 a^4} \ln \size {\frac {x^4 - a^4} {x^4} } + C$
directly.
== Sources ==
* : $\S 14$: Integrals involving $x^4 \pm a^4$: $14.322$
Category:Primitives involving x to the fourth plus or minus a to the fourth
Primitive of Reciprocal of x by x fourth plus a fourth

== Theorem ==
:$\ds \int \frac {\d x} {x \paren {x^4 + a^4} } = \frac 1 {4 a^4} \map \ln {\frac {x^4} {x^4 + a^4} }$ 
== Proof ==
From Primitive of Reciprocal of x by Power of x plus Power of a:
:$\ds \int \frac {\d x} {x \paren {x^n + a^n} } = \frac 1 {n a^n} \ln \size {\frac {x^n} {x^n + a^n} } + C$
So:
directly.
== Sources ==
* : $\S 14$: Integrals involving $x^4 \pm a^4$: $14.315$
Category:Primitives involving x to the fourth plus or minus a to the fourth
Primitive of Reciprocal of x by x squared minus a squared

== Theorem ==
:$\ds \int \frac {\d x} {x \paren {x^2 - a^2} } = \frac 1 {2 a^2} \map \ln {\frac {x^2 - a^2} {x^2} } + C$
for $x^2 > a^2$.
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {x \paren {a^2 - x^2} }$
== Sources ==
* : $\S 14$: Integrals involving $x^2 - a^2$, $x^2 > a^2$: $14.148$
* : $\S 17$: Tables of Special Indefinite Integrals: $(7)$ Integrals Involving $x^2 - a^2$, $x^2 > a^2$: $17.7.5.$
Category:Primitive of Reciprocal of x by x squared minus a squared
Category:Primitives involving x squared minus a squared
Primitive of Reciprocal of x by x squared minus a squared/Partial Fraction Expansion

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

== Theorem ==
:$\ds \int \frac {\d x} {x \paren {x^2 - a^2}^2} = \frac {-1} {2 a^2 \paren {x^2 - a^2} } + \frac 1 {2 a^4} \map \ln {\frac {x^2} {x^2 - a^2} } + C$
for $x^2 > a^2$.
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {x \paren {a^2 - x^2}^2}$
== Sources ==
* : $\S 14$: Integrals involving $x^2 - a^2$, $x^2 > a^2$: $14.155$
* : $\S 17$: Tables of Special Indefinite Integrals: $(7)$ Integrals Involving $x^2 - a^2$, $x^2 > a^2$: $17.7.12.$
Category:Primitive of Reciprocal of x by x squared minus a squared squared
Category:Primitives involving x squared minus a squared
Primitive of Reciprocal of x by x squared minus a squared squared/Partial Fraction Expansion

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

== Theorem ==
:$\ds \int \frac {\rd x} {x \paren {x^2 + a^2} } = \frac 1 {2 a^2} \map \ln {\frac {x^2} {x^2 + a^2} } + C$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : $\S 14$: Integrals involving $x^2 + a^2$: $14.129$
* : $\S 17$: Tables of Special Indefinite Integrals: $(6)$ Integrals Involving $x^2 + a^2$: $17.6.5.$
Category:Primitive of Reciprocal of x by x squared plus a squared
Category:Primitives involving x squared plus a squared
Primitive of Reciprocal of x by x squared plus a squared/Partial Fraction Expansion

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

== Theorem ==
== Proof ==
From Primitive of $\dfrac 1 {x \paren {x^n + a^n} }$:
:$\ds \int \frac {\d x} {x \paren {x^n + a^n} } = \frac 1 {n a^n} \ln \size {\frac {x^n} {x^n + a^n} } + C$
So:
Category:Primitive of Reciprocal of x by x squared plus a squared
Primitive of Reciprocal of x by x squared plus a squared squared

== Theorem ==
:$\ds \int \frac {\d x} {x \paren {x^2 + a^2}^2} = \frac 1 {2 a^2 \paren {x^2 + a^2} } + \frac 1 {2 a^4} \map \ln {\frac {x^2} {x^2 + a^2} } + C$
== Proof ==
Let:
== Sources ==
* : $\S 14$: Integrals involving $x^2 + a^2$: $14.136$
* : $\S 17$: Tables of Special Indefinite Integrals: $(6)$ Integrals Involving $x^2 + a^2$: $17.6.12.$
Category:Primitive of Reciprocal of x by x squared plus a squared squared
Category:Primitives involving x squared plus a squared
Primitive of Reciprocal of x by x squared plus a squared squared/Partial Fraction Expansion

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

== Theorem ==
:$\ds \int \frac {\d x} {x^3 \sqrt {a^2 - x^2} } = \frac {-\sqrt {a^2 - x^2} } {2 a^2 x^2} - \frac 1 {2 a^3} \map \ln {\frac {a + \sqrt {a^2 - x^2} } {\size x} } + C$
== 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 $\dfrac 1 {x^3 \sqrt {x^2 + a^2} }$
* Primitive of $\dfrac 1 {x^3 \sqrt {x^2 - a^2} }$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a^2 - x^2}$: $14.243$
Category:Primitives involving Root of a squared minus x squared
Primitive of Reciprocal of x cubed by Root of a squared minus x squared cubed

== Theorem ==
:$\ds \int \frac {\d x} {x^3 \paren {\sqrt {a^2 - x^2} }^3} = \frac {-1} {2 a^2 x^2 \sqrt {a^2 - x^2} } + \frac 3 {2 a^4 \sqrt {a^2 - x^2} } - \frac 3 {2 a^5} \map \ln {\frac {a + \sqrt {a^2 - x^2} } {\size x} } + C$
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {x^3 \paren {\sqrt {x^2 + a^2} }^3}$
* Primitive of $\dfrac 1 {x^3 \paren {\sqrt {x^2 - a^2} }^3}$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a^2 - x^2}$: $14.257$
Category:Primitives involving Root of a squared minus x squared
Primitive of Reciprocal of x cubed by Root of x squared minus a squared

== Theorem ==
:$\ds \int \frac {\d x} {x^3 \sqrt {x^2 - a^2} } = \frac {\sqrt {x^2 - a^2} } {2 a^2 x^2} + \frac 1 {2 a^3} \arcsec \size {\frac x a} + C$
for $\size x > a$.
== 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 Reciprocal of $x^3 \sqrt {x^2 + a^2}$
* Primitive of Reciprocal of $x^3 \sqrt {a^2 - x^2}$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {x^2 - a^2}$: $14.215$
Category:Primitives involving Root of x squared minus a squared
Primitive of Reciprocal of x cubed by Root of x squared minus a squared cubed

== Theorem ==
:$\ds \int \frac {\d x} {x^3 \paren {\sqrt {x^2 - a^2} }^3} = \frac 1 {2 a^2 x^2 \sqrt {x^2 - a^2} } - \frac 3 {2 a^4 \sqrt {x^2 - a^2} } + \frac 3 {2 a^5} \arcsec \size {\frac x a} + C$
for $\size x > a$.
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {x^3 \paren {\sqrt {x^2 + a^2} }^3}$
* Primitive of $\dfrac 1 {x^3 \paren {\sqrt {a^2 - x^2} }^3}$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {x^2 - a^2}$: $14.229$
Category:Primitives involving Root of x squared minus a squared
Primitive of Reciprocal of x cubed by Root of x squared plus a squared

== Theorem ==
:$\ds \int \frac {\d x} {x^3 \sqrt {x^2 + a^2} } = \frac {-\sqrt {x^2 + a^2} } {2 a^2 x^2} + \frac 1 {2 a^3} \map \ln {\frac {a + \sqrt {x^2 + a^2} } x} + C$
== 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 Reciprocal of $x^3 \sqrt {x^2 - a^2}$
* Primitive of Reciprocal of $x^3 \sqrt {a^2 - x^2}$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {x^2 + a^2}$: $14.188$
Category:Primitives involving Root of x squared plus a squared
Primitive of Reciprocal of x cubed by Root of x squared plus a squared cubed

== Theorem ==
:$\ds \int \frac {\d x} {x^3 \paren {\sqrt {x^2 + a^2} }^3} = \frac {-1} {2 a^2 x^2 \sqrt {x^2 + a^2} } - \frac 3 {2 a^4 \sqrt {x^2 + a^2} } + \frac 3 {2 a^5} \map \ln {\frac {a + \sqrt {x^2 + a^2} } x} + C$
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {x^3 \paren {\sqrt {x^2 - a^2} }^3}$
* Primitive of $\dfrac 1 {x^3 \paren {\sqrt {a^2 - x^2} }^3}$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {x^2 + a^2}$: $14.202$
Category:Primitives involving Root of x squared plus a squared
Primitive of Reciprocal of x cubed by a squared minus x squared

== Theorem ==
:$\ds \int \frac {\d x} {x^3 \paren {a^2 - x^2} } = \frac {-1} {2 a^2 x^2} + \frac 1 {2 a^4} \map \ln {\frac {x^2} {a^2 - x^2} } + C$
for $x^2 < a^2$.
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {x^3 \paren {x^2 - a^2} }$
== Sources ==
* : $\S 14$: Integrals involving $a^2 - x^2$, $x^2 < a^2$: $14.169$
Category:Primitives involving a squared minus x squared
Primitive of Reciprocal of x cubed by a squared minus x squared/Partial Fraction Expansion

== Lemma for Primitive of Reciprocal of $x^3 \paren {a^2 - x^2}$ ==
:$\dfrac 1 {x^3 \paren {a^2 - x^2} } \equiv \dfrac 1 {a^2 x^3} + \dfrac 1 {a^4 x} + \dfrac x {a^4 \paren {a^2 - x^2} }$
== 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:Primitives involving a squared minus x squared/Lemmata
Primitive of Reciprocal of x cubed by a squared minus x squared squared

== Theorem ==
:$\ds \int \frac {\d x} {x^3 \paren {a^2 - x^2}^2} = \frac {-1} {2 a^4 x^2} + \frac 1 {2 a^4 \paren {a^2 - x^2} } + \frac 1 {a^6} \map \ln {\frac {x^2} {a^2 - x^2} } + C$
for $x^2 < a^2$.
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {x^3 \paren {x^2 - a^2}^2}$
== Sources ==
* : $\S 14$: Integrals involving $a^2 - x^2$, $x^2 < a^2$: $14.176$
Category:Primitives involving a squared minus x squared
Primitive of Reciprocal of x cubed by a squared minus x squared squared/Partial Fraction Expansion

== Lemma for Primitive of Reciprocal of $x^3 \paren {a^2 - x^2}^2$ ==
:$\dfrac 1 {x^3 \paren {a^2 - x^2}^2} \equiv \dfrac 1 {a^4 x^3} + \dfrac 2 {a^6 x} + \dfrac {2 x} {a^6 \paren {a^2 - x^2} } + \dfrac x {a^4 \paren {a^2 - x^2}^2}$
== 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:Primitives involving a squared minus x squared/Lemmata
Primitive of Reciprocal of x cubed by a x + b

== Theorem ==
:$\ds \int \frac {\d x} {x^3 \paren {a x + b} } = \frac {2 a x - b} {2 b^2 x^2} + \frac {a^2} {b^3} \ln \size {\frac x {a x + b} } + C$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 14$: Integrals involving $a x + b$: $14.65$
Category:Primitive of Reciprocal of x cubed by a x + b‎
Category:Primitives involving a x + b
Category:Primitives involving Reciprocals
Primitive of Reciprocal of x cubed by a x + b/Partial Fraction Expansion

== Lemma for Primitive of $\dfrac 1 {x^3 \paren {a x + b} }$ ==
:$\dfrac 1 {x^3 \paren {a x + b} } \equiv \dfrac {a^2} {b^3 x} + \dfrac {-a} {b^2 x^2} + \dfrac 1 {b x^3} + \dfrac {-a^3} {b^3 \paren {a x + b} }$
== Proof ==
Setting $a x + b = 0$ in $(1)$:
Equating constants in $(1)$:
Equating $3$rd powers of $x$ 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 cubed by a x + b cubed

== Theorem ==
:$\ds \int \frac {\d x} {x^3 \paren {a x + b}^3} = \frac {a^4 x^2} {2 b^5 \paren {a x + b}^2} - \frac {4 a^3 x} {b^5 \paren {a x + b} } - \frac {\paren {a x + b}^2} {2 b^5 x^2} + \frac {4 a} {b^4 x} + \frac {6 a^2} {b^5} \ln \size {\frac x {a x + b} } + C$
== Proof ==
A partial fraction expansion of the integrand gives:
:$\dfrac 1 {x^3 \paren {a x + b}^3} = \dfrac {6 a^2} {b^5 x} - \dfrac {3 a} {b^4 x^2} + \dfrac 1 {b^3 x^3} - \dfrac {6 a^3} {b^5 \paren {a x + b} } - \dfrac {3 a^3} {b^4 \paren {a x + b}^2} - \dfrac {a^3} {b^3 \paren {a x + b}^3}$
From Linear Combination of Primitives it follows that the expression can be rendered as:
:$\ds \int \frac {\d x} {x^3 \paren {a x + b}^3} = \frac {6 a^2} {b^5} \int \frac {\d x} x - \frac {3 a} {b^4} \int \frac {\d x} {x^2} + \frac 1 {b^3} \int \frac {\d x} {x^3} - \frac {6 a^3} {b^5} \int \frac {\d x} {\paren {a x + b} } - \frac {3 a^3} {b^4} \int \frac {\d x} {\paren {a x + b}^2} - \frac {a^3} {b^3} \int \frac {\d x} {\paren {a x + b}^3}$
This can be integrated termwise as follows (ignoring for now the arbitrary constant):
The terms with the logarithm can be combined thus:
and so the solution to the integral can be expressed as:
While this would usually be considered as an acceptable form to leave such an expression, there is some way to go to obtain the result requested.
So, placing the non-logarithmic terms over a common denominator and rearranging, with a view to our destination:
Thus the full solution to the integral can be assembled as follows, where $\dfrac {4 a^2} {b^5}$ is subsumed into the arbitrary constant:
== Sources ==
* : $\S 14$: Integrals involving $a x + b$: $14.79$
::(in which a mistake apppears)
Category:Primitives involving a x + b‎
Primitive of Reciprocal of x cubed by a x + b cubed/Mistake

== Source Work ==
:Chapter $14$: Indefinite Integrals
::Integrals Involving $a x + b$: $14.79$
This mistake can be seen in the edition as published by Schaum: ISBN 0-07-060224-7 (unknown printing).
== Mistake ==
:$\ds \int \frac {\d x} {x^3 \paren {a x + b}^3} = \frac {a^4 x^2} {2 b^5 \paren {a x + b}^2} - \frac {4 a^3 x} {b^5 \paren {a x + b} } - \frac {\paren {a x + b}^2} {2 b^5 x^2} + \frac {6 a^2} {b^5} \ln \size {\frac x {a x + b} } + C$
== Correction ==
As demonstrated in Primitive of $\dfrac 1 {x^3 \paren {a x + b}^3}$ the correct expression is in fact:
:$\ds \int \frac {\d x} {x^3 \paren {a x + b}^3} = \frac {a^4 x^2} {2 b^5 \paren {a x + b}^2} - \frac {4 a^3 x} {b^5 \paren {a x + b} } - \frac {\paren {a x + b}^2} {2 b^5 x^2} + \frac {4 a} {b^4 x} + \frac {6 a^2} {b^5} \ln \size {\frac x {a x + b} } + C$
== Sources ==
* : $\S 14$: Integrals involving $a x + b$: $14.79$
Category:Mistakes/Mathematical Handbook of Formulas and Tables
Primitive of Reciprocal of x cubed by a x + b cubed/Partial Fraction Expansion

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

== Theorem ==
:$\ds \int \frac {\d x} {x^3 \paren {a x + b}^2} = - \frac {\paren {a x + b}^2} {2 b^4 x^2} + \frac {3 a \paren {a x + b} } {b^4 x} - \frac {a^3 x} {b^4 \paren {a x + b} } + \frac {3 a^2} {b^4} \ln \size {\frac x {a x + b} } + C$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $a x + b$: $14.72$
Category:Primitives involving a x + b‎
Category:Primitives involving Reciprocals
Primitive of Reciprocal of x cubed by a x + b squared/Partial Fraction Expansion

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

== Theorem ==
:$\ds \int \frac {\d x} {x^3 \paren {x^4 - a^4} } = \frac 1 {2 a^4 x^2} + \frac 1 {4 a^6} \ln \size {\frac {x^2 - a^2} {x^2 + a^2} } + C$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $x^4 \pm a^4$: $14.324$
* : $\S 17$: Tables of Special Indefinite Integrals: $(15)$ Integrals Involving $x^4 \pm a^4$: $17.15.14.$
Category:Primitives involving x to the fourth plus or minus a to the fourth
Primitive of Reciprocal of x cubed by x fourth plus a fourth

== Theorem ==
:$\ds \int \frac {\d x} {x^3 \paren {x^4 + a^4} } = \frac {-1} {2 a^4 x^2} - \frac 1 {2 a^6} \arctan \frac {x^2} {a^2}$ 
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $x^4 \pm a^4$: $14.317$
Category:Primitives involving x to the fourth plus or minus a to the fourth
Primitive of Reciprocal of x cubed by x squared minus a squared

== Theorem ==
:$\ds \int \frac {\d x} {x^3 \paren {x^2 - a^2} } = \frac 1 {2 a^2 x^2} + \frac 1 {2 a^4} \map \ln {\frac {x^2 - a^2} {x^2} } + C$
for $x^2 > a^2$.
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {x^3 \paren {a^2 - x^2} }$
== Sources ==
* : $\S 14$: Integrals involving $x^2 - a^2$, $x^2 > a^2$: $14.150$
* : $\S 17$: Tables of Special Indefinite Integrals: $(7)$ Integrals Involving $x^2 - a^2$, $x^2 > a^2$: $17.7.7.$
Category:Primitive of Reciprocal of x cubed by x squared minus a squared
Category:Primitives involving x squared minus a squared