Таблиця інтегралів ірраціональних функцій

Це список інтегралів (первісних функцій) ірраціональних функцій. Для повнішого списку інтегралів дивись таблицю інтегралів.

Інтеграли, що включають ${\displaystyle r={\sqrt {x^{2}+a^{2}$

${\displaystyle \int r\;dx={\frac {1}{2}\left(xr+a^{2}\,\ln \left(x+r\right)\right)}$

${\displaystyle \int r^{3}\;dx={\frac {1}{4}xr^{3}+{\frac {3}{8}a^{2}xr+{\frac {3}{8}a^{4}\ln \left(x+r\right)}$

${\displaystyle \int r^{5}\;dx={\frac {1}{6}xr^{5}+{\frac {5}{24}a^{2}xr^{3}+{\frac {5}{16}a^{4}xr+{\frac {5}{16}a^{6}\ln \left(x+r\right)}$

${\displaystyle \int xr\;dx={\frac {r^{3}{3}$

${\displaystyle \int xr^{3}\;dx={\frac {r^{5}{5}$

${\displaystyle \int xr^{2n+1}\;dx={\frac {r^{2n+3}{2n+3}$

${\displaystyle \int x^{2}r\;dx={\frac {xr^{3}{4}-{\frac {a^{2}xr}{8}-{\frac {a^{4}{8}\ln \left(x+r\right)}$

${\displaystyle \int x^{2}r^{3}\;dx={\frac {xr^{5}{6}-{\frac {a^{2}xr^{3}{24}-{\frac {a^{4}xr}{16}-{\frac {a^{6}{16}\ln \left(x+r\right)}$

${\displaystyle \int x^{3}r\;dx={\frac {r^{5}{5}-{\frac {a^{2}r^{3}{3}$

${\displaystyle \int x^{3}r^{3}\;dx={\frac {r^{7}{7}-{\frac {a^{2}r^{5}{5}$

${\displaystyle \int x^{3}r^{2n+1}\;dx={\frac {r^{2n+5}{2n+5}-{\frac {a^{3}r^{2n+3}{2n+3}$

${\displaystyle \int x^{4}r\;dx={\frac {x^{3}r^{3}{6}-{\frac {a^{2}xr^{3}{8}+{\frac {a^{4}xr}{16}+{\frac {a^{6}{16}\ln \left(x+r\right)}$

${\displaystyle \int x^{4}r^{3}\;dx={\frac {x^{3}r^{5}{8}-{\frac {a^{2}xr^{5}{16}+{\frac {a^{4}xr^{3}{64}+{\frac {3a^{6}xr}{128}+{\frac {3a^{8}{128}\ln \left(x+r\right)}$

${\displaystyle \int x^{5}r\;dx={\frac {r^{7}{7}-{\frac {2a^{2}r^{5}{5}+{\frac {a^{4}r^{3}{3}$

${\displaystyle \int x^{5}r^{3}\;dx={\frac {r^{9}{9}-{\frac {2a^{2}r^{7}{7}+{\frac {a^{4}r^{5}{5}$

${\displaystyle \int x^{5}r^{2n+1}\;dx={\frac {r^{2n+7}{2n+7}-{\frac {2a^{2}r^{2n+5}{2n+5}+{\frac {a^{4}r^{2n+3}{2n+3}$

${\displaystyle \int {\frac {r\;dx}{x}=r-a\ln \left|{\frac {a+r}{x}\right|=r-a\,\operatorname {arsinh} {\frac {a}{x}$

${\displaystyle \int {\frac {r^{3}\;dx}{x}={\frac {r^{3}{3}+a^{2}r-a^{3}\ln \left|{\frac {a+r}{x}\right|}$

${\displaystyle \int {\frac {r^{5}\;dx}{x}={\frac {r^{5}{5}+{\frac {a^{2}r^{3}{3}+a^{4}r-a^{5}\ln \left|{\frac {a+r}{x}\right|}$

${\displaystyle \int {\frac {r^{7}\;dx}{x}={\frac {r^{7}{7}+{\frac {a^{2}r^{5}{5}+{\frac {a^{4}r^{3}{3}+a^{6}r-a^{7}\ln \left|{\frac {a+r}{x}\right|}$

${\displaystyle \int {\frac {dx}{r}=\operatorname {arsinh} {\frac {x}{a}=\ln \left({\frac {x+r}{a}\right)}$

${\displaystyle \int {\frac {dx}{r^{3}={\frac {x}{a^{2}r}$

${\displaystyle \int {\frac {x\,dx}{r}=r}$

${\displaystyle \int {\frac {x\,dx}{r^{3}=-{\frac {1}{r}$

${\displaystyle \int {\frac {x^{2}\;dx}{r}={\frac {x}{2}r-{\frac {a^{2}{2}\,\operatorname {arsinh} {\frac {x}{a}={\frac {x}{2}r-{\frac {a^{2}{2}\ln \left({\frac {x+r}{a}\right)}$

${\displaystyle \int {\frac {dx}{xr}=-{\frac {1}{a}\,\operatorname {arsinh} {\frac {a}{x}=-{\frac {1}{a}\ln \left|{\frac {a+r}{x}\right|}$

Інтеграли, що включають${\displaystyle s={\sqrt {x^{2}-a^{2}$:
Припустимо ${\displaystyle (x^{2}>a^{2})}$, для ${\displaystyle (x^{2}<a^{2})}$, дивись наступну секцію:

${\displaystyle \int s\;dx={\frac {1}{2}\left(xs-a^{2}\ln(x+s)\right)}$

${\displaystyle \int xs\;dx={\frac {1}{3}s^{3}$

${\displaystyle \int {\frac {s\;dx}{x}=s-a\arccos \left|{\frac {a}{x}\right|}$

${\displaystyle \int {\frac {dx}{s}=\int {\frac {dx}{\sqrt {x^{2}-a^{2}=\ln \left|{\frac {x+s}{a}\right|}$

Зауважте, що ${\displaystyle \ln \left|{\frac {x+s}{a}\right|=\mathrm {sgn} (x)\,\operatorname {arcosh} \left|{\frac {x}{a}\right|={\frac {1}{2}\ln \left({\frac {x+s}{x-s}\right)}$, where the positive value of ${\displaystyle \operatorname {arcosh} \left|{\frac {x}{a}\right|}$ is to be taken.

${\displaystyle \int {\frac {x\;dx}{s}=s}$

${\displaystyle \int {\frac {x\;dx}{s^{3}=-{\frac {1}{s}$

${\displaystyle \int {\frac {x\;dx}{s^{5}=-{\frac {1}{3s^{3}$

${\displaystyle \int {\frac {x\;dx}{s^{7}=-{\frac {1}{5s^{5}$

${\displaystyle \int {\frac {x\;dx}{s^{2n+1}=-{\frac {1}{(2n-1)s^{2n-1}$

${\displaystyle \int {\frac {x^{2m}\;dx}{s^{2n+1}=-{\frac {1}{2n-1}{\frac {x^{2m-1}{s^{2n-1}+{\frac {2m-1}{2n-1}\int {\frac {x^{2m-2}\;dx}{s^{2n-1}$

${\displaystyle \int {\frac {x^{2}\;dx}{s}={\frac {xs}{2}+{\frac {a^{2}{2}\ln \left|{\frac {x+s}{a}\right|}$

${\displaystyle \int {\frac {x^{2}\;dx}{s^{3}=-{\frac {x}{s}+\ln \left|{\frac {x+s}{a}\right|}$

${\displaystyle \int {\frac {x^{4}\;dx}{s}={\frac {x^{3}s}{4}+{\frac {3}{8}a^{2}xs+{\frac {3}{8}a^{4}\ln \left|{\frac {x+s}{a}\right|}$

${\displaystyle \int {\frac {x^{4}\;dx}{s^{3}={\frac {xs}{2}-{\frac {a^{2}x}{s}+{\frac {3}{2}a^{2}\ln \left|{\frac {x+s}{a}\right|}$

${\displaystyle \int {\frac {x^{4}\;dx}{s^{5}=-{\frac {x}{s}-{\frac {1}{3}{\frac {x^{3}{s^{3}+\ln \left|{\frac {x+s}{a}\right|}$

${\displaystyle \int {\frac {x^{2m}\;dx}{s^{2n+1}=(-1)^{n-m}{\frac {1}{a^{2(n-m)}\sum _{i=0}^{n-m-1}{\frac {1}{2(m+i)+1}{n-m-1 \choose i}{\frac {x^{2(m+i)+1}{s^{2(m+i)+1}\qquad {\mbox{(}n>m\geq 0{\mbox{)}$

${\displaystyle \int {\frac {dx}{s^{3}=-{\frac {1}{a^{2}{\frac {x}{s}$

${\displaystyle \int {\frac {dx}{s^{5}={\frac {1}{a^{4}\left[{\frac {x}{s}-{\frac {1}{3}{\frac {x^{3}{s^{3}\right]}$

${\displaystyle \int {\frac {dx}{s^{7}=-{\frac {1}{a^{6}\left[{\frac {x}{s}-{\frac {2}{3}{\frac {x^{3}{s^{3}+{\frac {1}{5}{\frac {x^{5}{s^{5}\right]}$

${\displaystyle \int {\frac {dx}{s^{9}={\frac {1}{a^{8}\left[{\frac {x}{s}-{\frac {3}{3}{\frac {x^{3}{s^{3}+{\frac {3}{5}{\frac {x^{5}{s^{5}-{\frac {1}{7}{\frac {x^{7}{s^{7}\right]}$

${\displaystyle \int {\frac {x^{2}\;dx}{s^{5}=-{\frac {1}{a^{2}{\frac {x^{3}{3s^{3}$

${\displaystyle \int {\frac {x^{2}\;dx}{s^{7}={\frac {1}{a^{4}\left[{\frac {1}{3}{\frac {x^{3}{s^{3}-{\frac {1}{5}{\frac {x^{5}{s^{5}\right]}$

${\displaystyle \int {\frac {x^{2}\;dx}{s^{9}=-{\frac {1}{a^{6}\left[{\frac {1}{3}{\frac {x^{3}{s^{3}-{\frac {2}{5}{\frac {x^{5}{s^{5}+{\frac {1}{7}{\frac {x^{7}{s^{7}\right]}$

Інтеграли, що включають ${\displaystyle u={\sqrt {a^{2}-x^{2}$:

${\displaystyle \int u\;dx={\frac {1}{2}\left(xu+a^{2}\arcsin {\frac {x}{a}\right)\qquad {\mbox{(}|x|\leq |a|{\mbox{)}$

${\displaystyle \int xu\;dx=-{\frac {1}{3}u^{3}\qquad {\mbox{(}|x|\leq |a|{\mbox{)}$

${\displaystyle \int x^{2}u\;dx=-{\frac {x}{4}u^{3}+{\frac {a^{2}{8}(xu+a^{2}\arcsin {\frac {x}{a})\qquad {\mbox{(}|x|\leq |a|{\mbox{)}$

${\displaystyle \int {\frac {u\;dx}{x}=u-a\ln \left|{\frac {a+u}{x}\right|\qquad {\mbox{(}|x|\leq |a|{\mbox{)}$

${\displaystyle \int {\frac {dx}{u}=\arcsin {\frac {x}{a}\qquad {\mbox{(}|x|\leq |a|{\mbox{)}$

${\displaystyle \int {\frac {x^{2}\;dx}{u}={\frac {1}{2}\left(-xu+a^{2}\arcsin {\frac {x}{a}\right)\qquad {\mbox{(}|x|\leq |a|{\mbox{)}$

${\displaystyle \int u\;dx={\frac {1}{2}\left(xu-\operatorname {sgn} x\,\operatorname {arcosh} \left|{\frac {x}{a}\right|\right)\qquad {\mbox{(for }|x|\geq |a|{\mbox{)}$

${\displaystyle \int {\frac {x}{u}\;dx=-u\qquad {\mbox{(}|x|\leq |a|{\mbox{)}$

Інтеграли, що включають ${\displaystyle S={\sqrt {ax+b}$:

${\displaystyle \int S{dx}={\frac {2S^{3}{3a}$

${\displaystyle \int {\frac {dx}{S}={\frac {2S}{a}$

${\displaystyle \int {\frac {dx}{xS}={\begin{cases}-{\frac {2}{\sqrt {b}\mathrm {arcoth} \left({\frac {S}{\sqrt {b}\right)&{\mbox{(for }b>0,\quad ax>0{\mbox{)}\\-{\frac {2}{\sqrt {b}\mathrm {artanh} \left({\frac {S}{\sqrt {b}\right)&{\mbox{(for }b>0,\quad ax<0{\mbox{)}\\{\frac {2}{\sqrt {-b}\arctan \left({\frac {S}{\sqrt {-b}\right)&{\mbox{(for }b<0{\mbox{)}\\\end{cases}$

${\displaystyle \int {\frac {S}{x}\,dx={\begin{cases}2\left(S-{\sqrt {b}\,\mathrm {arcoth} \left({\frac {S}{\sqrt {b}\right)\right)&{\mbox{(for }b>0,\quad ax>0{\mbox{)}\\2\left(S-{\sqrt {b}\,\mathrm {artanh} \left({\frac {S}{\sqrt {b}\right)\right)&{\mbox{(for }b>0,\quad ax<0{\mbox{)}\\2\left(S-{\sqrt {-b}\arctan \left({\frac {S}{\sqrt {-b}\right)\right)&{\mbox{(for }b<0{\mbox{)}\\\end{cases}$

${\displaystyle \int {\frac {x^{n}{S}dx={\frac {2}{a(2n+1)}\left(x^{n}S-bn\int {\frac {x^{n-1}{S}dx\right)}$

${\displaystyle \int x^{n}Sdx={\frac {2}{a(2n+3)}\left(x^{n}S^{3}-nb\int x^{n-1}Sdx\right)}$

${\displaystyle \int {\frac {1}{x^{n}S}dx=-{\frac {1}{b(n-1)}\left({\frac {S}{x^{n-1}+\left(n-{\frac {3}{2}\right)a\int {\frac {dx}{x^{n-1}S}\right)}$

Інтеграли, що включають ${\displaystyle R={\sqrt {ax^{2}+bx+c}$:

Нехай (ax² + bx + c) не можна привести до вигляду (px + q)² для деяких p та q, тоді:

${\displaystyle \int {\frac {dx}{R}={\frac {1}{\sqrt {a}\ln \left|2{\sqrt {a}R+2ax+b\right|\qquad {\mbox{(for }a>0{\mbox{)}$

${\displaystyle \int {\frac {dx}{R}={\frac {1}{\sqrt {a}\,\operatorname {arsinh} {\frac {2ax+b}{\sqrt {4ac-b^{2}\qquad {\mbox{(for }a>0{\mbox{, }4ac-b^{2}>0{\mbox{)}$

${\displaystyle \int {\frac {dx}{R}={\frac {1}{\sqrt {a}\ln |2ax+b|\quad {\mbox{(for }a>0{\mbox{, }4ac-b^{2}=0{\mbox{)}$

${\displaystyle \int {\frac {dx}{R}=-{\frac {1}{\sqrt {-a}\arcsin {\frac {2ax+b}{\sqrt {b^{2}-4ac}\qquad {\mbox{(for }a<0{\mbox{, }4ac-b^{2}<0{\mbox{, }\left|2ax+b\right|<{\sqrt {b^{2}-4ac}{\mbox{)}$

${\displaystyle \int {\frac {dx}{R^{3}={\frac {4ax+2b}{(4ac-b^{2})R}$

${\displaystyle \int {\frac {dx}{R^{5}={\frac {4ax+2b}{3(4ac-b^{2})R}\left({\frac {1}{R^{2}+{\frac {8a}{4ac-b^{2}\right)}$

${\displaystyle \int {\frac {dx}{R^{2n+1}={\frac {2}{(2n-1)(4ac-b^{2})}\left({\frac {2ax+b}{R^{2n-1}+4a(n-1)\int {\frac {dx}{R^{2n-1}\right)}$

${\displaystyle \int {\frac {x}{R}\;dx={\frac {R}{a}-{\frac {b}{2a}\int {\frac {dx}{R}$

${\displaystyle \int {\frac {x}{R^{3}\;dx=-{\frac {2bx+4c}{(4ac-b^{2})R}$

${\displaystyle \int {\frac {x}{R^{2n+1}\;dx=-{\frac {1}{(2n-1)aR^{2n-1}-{\frac {b}{2a}\int {\frac {dx}{R^{2n+1}$

${\displaystyle \int {\frac {dx}{xR}=-{\frac {1}{\sqrt {c}\ln \left({\frac {2{\sqrt {c}R+bx+2c}{x}\right)}$

${\displaystyle \int {\frac {dx}{xR}=-{\frac {1}{\sqrt {c}\operatorname {arsinh} \left({\frac {bx+2c}{|x|{\sqrt {4ac-b^{2}\right)}$

Інтеграли, що включають ${\displaystyle Y=(ax^{2}+bx+c)^{\frac {2m+1}{2}$, типу:

${\displaystyle \int {dx \over (ax^{2}+bx+c)^{\frac {2m+1}{2}=\left({4 \over 4ac-b^{2}\right)^{m}\int (a-t^{2})^{m-1}dt,}$

де ${\displaystyle m\!}$ є цілим додатнім числом, а змінна ${\displaystyle t\!}$ задається виразом ${\displaystyle t={ax+{\frac {b}{2} \over {\sqrt {ax^{2}+bx+c}\!}$ (див. Підстановка Абеля).

Для випадку ${\displaystyle m=1\!}$ матимемо:

${\displaystyle \int {dx \over (ax^{2}+bx+c)^{\frac {3}{2}={4 \over 4ac-b^{2}\int dt={4t \over {4ac-b^{2}=\left({4 \over 4ac-b^{2}\right)\;\;{ax+{\frac {b}{2} \over {\sqrt {ax^{2}+bx+c}.}$

Для випадку ${\displaystyle m=2\!}$ матимемо:

${\displaystyle \int {dx \over (ax^{2}+bx+c)^{\frac {5}{2}=\left({4 \over 4ac-b^{2}\right)^{2}\int (a-t^{2})dt=\left({4 \over 4ac-b^{2}\right)^{2}\left(at-{\frac {t^{3}{3}\right),}$

куди потім можна підставити явне значення для ${\displaystyle t\!}$ й спростити результат.

Для випадку ${\displaystyle m=3\!}$ матимемо:

${\displaystyle \int {dx \over (ax^{2}+bx+c)^{\frac {7}{2}=\left({4 \over 4ac-b^{2}\right)^{3}\int (a-t^{2})^{2}dt=\left({4 \over 4ac-b^{2}\right)^{3}\left(a^{2}t-{\frac {2a}{3}t^{3}+{\frac {t^{5}{5}\right)}$

і так далі.

• Abramowitz, Milton; Stegun, Irene Ann, ред. (1983). Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. Applied Mathematics Series. Т. 55 (вид. 9th). Washington D.C.; New York: United States Department of Commerce, National Bureau of Standards; Dover Publications. ISBN 0-486-61272-4. LCCN 64-60036. MR 0167642. ISBN 978-0-486-61272-0. LCCN 6512253-{3}. {cite book}: Зовнішнє посилання в |title= (довідка)
• Григорій Михайлович Фіхтенгольц. Курс диференціального та інтегрального числення. — 2025. — 2391 с.(укр.)
• Двайт Г. Б. Иррациональные алгебраические функции — интегралы // Таблицы интегралов и другие математические формулы / пер. с англ. Н. В. Леви ; под ред. К. А. Семендяева. — М. : Наука, 1978. — С. 40-69. (рос.)