Следващата таблица е на интеграли (примитивни функции) на ирационални функции. За по-пълна таблица на интеграли виж Таблични интеграли. За информацията в тази статия или раздел не са посочени източници. Въпросната информация може да е непълна, неточна или изцяло невярна.Имайте предвид, че това може да стане причина за изтриването на цялата статия или раздел. Интеграли, съдържащи r = x 2 + a 2 {\displaystyle r={\sqrt {x^{2}+a^{2}}}} ∫ r d x = 1 2 ( x r + a 2 ln ( x + r ) ) {\displaystyle \int r\;dx={\frac {1}{2}}\left(xr+a^{2}\,\ln \left(x+r\right)\right)} ∫ r 3 d x = 1 4 x r 3 + 1 8 3 a 2 x r + 3 8 a 4 ln ( x + r ) {\displaystyle \int r^{3}\;dx={\frac {1}{4}}xr^{3}+{\frac {1}{8}}3a^{2}xr+{\frac {3}{8}}a^{4}\ln \left(x+r\right)} ∫ r 5 d x = 1 6 x r 5 + 5 24 a 2 x r 3 + 5 16 a 4 x r + 5 16 a 6 ln ( x + r ) {\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)} ∫ x r d x = r 3 3 {\displaystyle \int xr\;dx={\frac {r^{3}}{3}}} ∫ x r 3 d x = r 5 5 {\displaystyle \int xr^{3}\;dx={\frac {r^{5}}{5}}} ∫ x r 2 n + 1 d x = r 2 n + 3 2 n + 3 {\displaystyle \int xr^{2n+1}\;dx={\frac {r^{2n+3}}{2n+3}}} ∫ x 2 r d x = x r 3 4 − a 2 x r 8 − a 4 8 ln ( x + r ) {\displaystyle \int x^{2}r\;dx={\frac {xr^{3}}{4}}-{\frac {a^{2}xr}{8}}-{\frac {a^{4}}{8}}\ln \left(x+r\right)} ∫ x 2 r 3 d x = x r 5 6 − a 2 x r 3 24 − a 4 x r 16 − a 6 16 ln ( x + r ) {\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)} ∫ x 3 r d x = r 5 5 − a 2 r 3 3 {\displaystyle \int x^{3}r\;dx={\frac {r^{5}}{5}}-{\frac {a^{2}r^{3}}{3}}} ∫ x 3 r 3 d x = r 7 7 − a 2 r 5 5 {\displaystyle \int x^{3}r^{3}\;dx={\frac {r^{7}}{7}}-{\frac {a^{2}r^{5}}{5}}} ∫ x 3 r 2 n + 1 d x = r 2 n + 5 2 n + 5 − a 3 r 2 n + 3 2 n + 3 {\displaystyle \int x^{3}r^{2n+1}\;dx={\frac {r^{2n+5}}{2n+5}}-{\frac {a^{3}r^{2n+3}}{2n+3}}} ∫ x 4 r d x = x 3 r 3 6 − a 2 x r 3 8 + a 4 x r 16 + a 6 16 ln ( x + r ) {\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)} ∫ x 4 r 3 d x = x 3 r 5 8 − a 2 x r 5 16 + a 4 x r 3 64 + 3 a 6 x r 128 + 3 a 8 128 ln ( x + r ) {\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)} ∫ x 5 r d x = r 7 7 − 2 a 2 r 5 5 + a 4 r 3 3 {\displaystyle \int x^{5}r\;dx={\frac {r^{7}}{7}}-{\frac {2a^{2}r^{5}}{5}}+{\frac {a^{4}r^{3}}{3}}} ∫ x 5 r 3 d x = r 9 9 − 2 a 2 r 7 7 + a 4 r 5 5 {\displaystyle \int x^{5}r^{3}\;dx={\frac {r^{9}}{9}}-{\frac {2a^{2}r^{7}}{7}}+{\frac {a^{4}r^{5}}{5}}} ∫ x 5 r 2 n + 1 d x = r 2 n + 7 2 n + 7 − 2 a 2 r 2 n + 5 2 n + 5 + a 4 r 2 n + 3 2 n + 3 {\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}}} ∫ r d x x = r − a ln | a + r x | = r − a sinh − 1 a x {\displaystyle \int {\frac {r\;dx}{x}}=r-a\ln \left|{\frac {a+r}{x}}\right|=r-a\sinh ^{-1}{\frac {a}{x}}} ∫ r 3 d x x = r 3 3 + a 2 r − a 3 ln | a + r x | {\displaystyle \int {\frac {r^{3}\;dx}{x}}={\frac {r^{3}}{3}}+a^{2}r-a^{3}\ln \left|{\frac {a+r}{x}}\right|} ∫ r 5 d x x = r 5 5 + a 2 r 3 3 + a 4 r − a 5 ln | a + r x | {\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|} ∫ r 7 d x x = r 7 7 + a 2 r 5 5 + a 4 r 3 3 + a 6 r − a 7 ln | a + r x | {\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|} ∫ d x r = sinh − 1 x a = ln | x + r | {\displaystyle \int {\frac {dx}{r}}=\sinh ^{-1}{\frac {x}{a}}=\ln \left|x+r\right|} ∫ d x r 3 = x a 2 r {\displaystyle \int {\frac {dx}{r^{3}}}={\frac {x}{a^{2}r}}} ∫ x d x r = r {\displaystyle \int {\frac {x\,dx}{r}}=r} ∫ x d x r 3 = − 1 r {\displaystyle \int {\frac {x\,dx}{r^{3}}}=-{\frac {1}{r}}} ∫ x 2 d x r = x 2 r − a 2 2 sinh − 1 x a = x 2 r − a 2 2 ln | x + r | {\displaystyle \int {\frac {x^{2}\;dx}{r}}={\frac {x}{2}}r-{\frac {a^{2}}{2}}\,\sinh ^{-1}{\frac {x}{a}}={\frac {x}{2}}r-{\frac {a^{2}}{2}}\ln \left|x+r\right|} ∫ d x x r = − 1 a sinh − 1 a x = − 1 a ln | a + r x | {\displaystyle \int {\frac {dx}{xr}}=-{\frac {1}{a}}\,\sinh ^{-1}{\frac {a}{x}}=-{\frac {1}{a}}\ln \left|{\frac {a+r}{x}}\right|} Remove adsИнтеграли, съдържащи s = x 2 − a 2 {\displaystyle s={\sqrt {x^{2}-a^{2}}}} За случаите при които ( x 2 > a 2 ) {\displaystyle (x^{2}>a^{2})} , за случаите ( x 2 < a 2 ) {\displaystyle (x^{2}<a^{2})} , виж следващата секция: ∫ x s d x = 1 3 s 3 {\displaystyle \int xs\;dx={\frac {1}{3}}s^{3}} ∫ s d x x = s − a cos − 1 | a x | {\displaystyle \int {\frac {s\;dx}{x}}=s-a\cos ^{-1}\left|{\frac {a}{x}}\right|} ∫ d x s = ∫ d x x 2 − a 2 = ln | x + s a | {\displaystyle \int {\frac {dx}{s}}=\int {\frac {dx}{\sqrt {x^{2}-a^{2}}}}=\ln \left|{\frac {x+s}{a}}\right|} Забележка: При ln | x + s a | = s g n ( x ) cosh − 1 | x a | = 1 2 ln ( x + s x − s ) {\displaystyle \ln \left|{\frac {x+s}{a}}\right|=\mathrm {sgn} (x)\cosh ^{-1}\left|{\frac {x}{a}}\right|={\frac {1}{2}}\ln \left({\frac {x+s}{x-s}}\right)} , трябва да се вземе само положителната стойност на cosh − 1 | x a | {\displaystyle \cosh ^{-1}\left|{\frac {x}{a}}\right|} . ∫ x d x s = s {\displaystyle \int {\frac {x\;dx}{s}}=s} ∫ x d x s 3 = − 1 s {\displaystyle \int {\frac {x\;dx}{s^{3}}}=-{\frac {1}{s}}} ∫ x d x s 5 = − 1 3 s 3 {\displaystyle \int {\frac {x\;dx}{s^{5}}}=-{\frac {1}{3s^{3}}}} ∫ x d x s 7 = − 1 5 s 5 {\displaystyle \int {\frac {x\;dx}{s^{7}}}=-{\frac {1}{5s^{5}}}} ∫ x d x s 2 n + 1 = − 1 ( 2 n − 1 ) s 2 n − 1 {\displaystyle \int {\frac {x\;dx}{s^{2n+1}}}=-{\frac {1}{(2n-1)s^{2n-1}}}} ∫ x 2 m d x s 2 n + 1 = − 1 2 n − 1 x 2 m − 1 s 2 n − 1 + 2 m − 1 2 n − 1 ∫ x 2 m − 2 d x s 2 n − 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}}}} ∫ x 2 d x s = x s 2 + a 2 2 ln | x + s a | {\displaystyle \int {\frac {x^{2}\;dx}{s}}={\frac {xs}{2}}+{\frac {a^{2}}{2}}\ln \left|{\frac {x+s}{a}}\right|} ∫ x 2 d x s 3 = − x s + ln | x + s a | {\displaystyle \int {\frac {x^{2}\;dx}{s^{3}}}=-{\frac {x}{s}}+\ln \left|{\frac {x+s}{a}}\right|} ∫ x 4 d x s = x 3 s 4 + 3 8 a 2 x s + 3 8 a 4 ln | x + s a | {\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|} ∫ x 4 d x s 3 = x s 2 − a 2 x s + 3 2 a 2 ln | x + s a | {\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|} ∫ x 4 d x s 5 = − x s − 1 3 x 3 s 3 + ln | x + s a | {\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|} ∫ x 2 m d x s 2 n + 1 = ( − 1 ) n − m 1 a 2 ( n − m ) ∑ i = 0 n − m − 1 1 2 ( m + i ) + 1 ( n − m − 1 i ) x 2 ( m + i ) + 1 s 2 ( m + i ) + 1 ( n > m ≥ 0 ) {\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{)}}} ∫ d x s 3 = − 1 a 2 x s {\displaystyle \int {\frac {dx}{s^{3}}}=-{\frac {1}{a^{2}}}{\frac {x}{s}}} ∫ d x s 5 = 1 a 4 [ x s − 1 3 x 3 s 3 ] {\displaystyle \int {\frac {dx}{s^{5}}}={\frac {1}{a^{4}}}\left[{\frac {x}{s}}-{\frac {1}{3}}{\frac {x^{3}}{s^{3}}}\right]} ∫ d x s 7 = − 1 a 6 [ x s − 2 3 x 3 s 3 + 1 5 x 5 s 5 ] {\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]} ∫ d x s 9 = 1 a 8 [ x s − 3 3 x 3 s 3 + 3 5 x 5 s 5 − 1 7 x 7 s 7 ] {\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]} ∫ x 2 d x s 5 = − 1 a 2 x 3 3 s 3 {\displaystyle \int {\frac {x^{2}\;dx}{s^{5}}}=-{\frac {1}{a^{2}}}{\frac {x^{3}}{3s^{3}}}} ∫ x 2 d x s 7 = 1 a 4 [ 1 3 x 3 s 3 − 1 5 x 5 s 5 ] {\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]} ∫ x 2 d x s 9 = − 1 a 6 [ 1 3 x 3 s 3 − 2 5 x 5 s 5 + 1 7 x 7 s 7 ] {\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]} Remove adsИнтеграли, съдържащи t = a 2 − x 2 {\displaystyle t={\sqrt {a^{2}-x^{2}}}} ∫ t d x = 1 2 ( x t + a 2 arcsin x a ) ( | x | ≤ | a | ) {\displaystyle \int t\;dx={\frac {1}{2}}\left(xt+a^{2}\arcsin {\frac {x}{a}}\right)\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}} ∫ x t d x = − 1 3 t 3 ( | x | ≤ | a | ) {\displaystyle \int xt\;dx=-{\frac {1}{3}}t^{3}\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}} ∫ t d x x = t − a ln | a + t x | ( | x | ≤ | a | ) {\displaystyle \int {\frac {t\;dx}{x}}=t-a\ln \left|{\frac {a+t}{x}}\right|\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}} ∫ d x t = arcsin x a ( | x | ≤ | a | ) {\displaystyle \int {\frac {dx}{t}}=\arcsin {\frac {x}{a}}\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}} ∫ x 2 d x t = 1 2 ( − x t + a 2 arcsin x a ) ( | x | ≤ | a | ) {\displaystyle \int {\frac {x^{2}\;dx}{t}}={\frac {1}{2}}\left(-xt+a^{2}\arcsin {\frac {x}{a}}\right)\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}} ∫ t d x = 1 2 ( x t − sgn x cosh − 1 | x a | ) (for | x | ≥ | a | ) {\displaystyle \int t\;dx={\frac {1}{2}}\left(xt-\operatorname {sgn} x\,\cosh ^{-1}\left|{\frac {x}{a}}\right|\right)\qquad {\mbox{(for }}|x|\geq |a|{\mbox{)}}} Remove adsИнтеграли, съдържащи R = a x 2 + b x + c {\displaystyle R={\sqrt {ax^{2}+bx+c}}} ∫ d x R = 1 a ln | 2 a R + 2 a x + b | (for a > 0 ) {\displaystyle \int {\frac {dx}{R}}={\frac {1}{\sqrt {a}}}\ln \left|2{\sqrt {a}}R+2ax+b\right|\qquad {\mbox{(for }}a>0{\mbox{)}}} ∫ d x R = 1 a sinh − 1 2 a x + b 4 a c − b 2 (for a > 0 , 4 a c − b 2 > 0 ) {\displaystyle \int {\frac {dx}{R}}={\frac {1}{\sqrt {a}}}\,\sinh ^{-1}{\frac {2ax+b}{\sqrt {4ac-b^{2}}}}\qquad {\mbox{(for }}a>0{\mbox{, }}4ac-b^{2}>0{\mbox{)}}} ∫ d x R = 1 a ln | 2 a x + b | (for a > 0 , 4 a c − b 2 = 0 ) {\displaystyle \int {\frac {dx}{R}}={\frac {1}{\sqrt {a}}}\ln |2ax+b|\quad {\mbox{(for }}a>0{\mbox{, }}4ac-b^{2}=0{\mbox{)}}} ∫ d x R = − 1 − a arcsin 2 a x + b b 2 − 4 a c (for a < 0 , 4 a c − b 2 < 0 , | 2 a x + b | < b 2 − 4 a c ) {\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{)}}} ∫ d x R 3 = 4 a x + 2 b ( 4 a c − b 2 ) R {\displaystyle \int {\frac {dx}{R^{3}}}={\frac {4ax+2b}{(4ac-b^{2})R}}} ∫ d x R 5 = 4 a x + 2 b 3 ( 4 a c − b 2 ) R ( 1 R 2 + 8 a 4 a c − b 2 ) {\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)} ∫ d x R 2 n + 1 = 2 ( 2 n − 1 ) ( 4 a c − b 2 ) ( 2 a x + b R 2 n − 1 + 4 a ( n − 1 ) ∫ d x R 2 n − 1 ) {\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)} ∫ x R d x = R a − b 2 a ∫ d x R {\displaystyle \int {\frac {x}{R}}\;dx={\frac {R}{a}}-{\frac {b}{2a}}\int {\frac {dx}{R}}} ∫ x R 3 d x = − 2 b x + 4 c ( 4 a c − b 2 ) R {\displaystyle \int {\frac {x}{R^{3}}}\;dx=-{\frac {2bx+4c}{(4ac-b^{2})R}}} ∫ x R 2 n + 1 d x = − 1 ( 2 n − 1 ) a R 2 n − 1 − b 2 a ∫ d x R 2 n + 1 {\displaystyle \int {\frac {x}{R^{2n+1}}}\;dx=-{\frac {1}{(2n-1)aR^{2n-1}}}-{\frac {b}{2a}}\int {\frac {dx}{R^{2n+1}}}} ∫ d x x R = − 1 c ln ( 2 c R + b x + 2 c x ) {\displaystyle \int {\frac {dx}{xR}}=-{\frac {1}{\sqrt {c}}}\ln \left({\frac {2{\sqrt {c}}R+bx+2c}{x}}\right)} ∫ d x x R = − 1 c sinh − 1 ( b x + 2 c | x | 4 a c − b 2 ) {\displaystyle \int {\frac {dx}{xR}}=-{\frac {1}{\sqrt {c}}}\sinh ^{-1}\left({\frac {bx+2c}{|x|{\sqrt {4ac-b^{2}}}}}\right)} Remove adsИнтеграли, съдържащи S = a x + b {\displaystyle S={\sqrt {ax+b}}} ∫ d x x a x + b = − 2 b tanh − 1 a x + b b {\displaystyle \int {\frac {dx}{x{\sqrt {ax+b}}}}\,=\,{\frac {-2}{\sqrt {b}}}\tanh ^{-1}{\sqrt {\frac {ax+b}{b}}}} ∫ a x + b x d x = 2 ( a x + b − b tanh − 1 a x + b b ) {\displaystyle \int {\frac {\sqrt {ax+b}}{x}}\,dx\;=\;2\left({\sqrt {ax+b}}-{\sqrt {b}}\tanh ^{-1}{\sqrt {\frac {ax+b}{b}}}\right)} ∫ x n a x + b d x = 2 a ( 2 n + 1 ) ( x n a x + b − b n ∫ x n − 1 a x + b d x ) {\displaystyle \int {\frac {x^{n}}{\sqrt {ax+b}}}\,dx\;=\;{\frac {2}{a(2n+1)}}\left(x^{n}{\sqrt {ax+b}}-bn\int {\frac {x^{n-1}}{\sqrt {ax+b}}}\,dx\right)} ∫ x n a x + b d x = 2 2 n + 1 ( x n + 1 a x + b + b x n a x + b − n b ∫ x n − 1 a x + b d x ) {\displaystyle \int x^{n}{\sqrt {ax+b}}\,dx\;=\;{\frac {2}{2n+1}}\left(x^{n+1}{\sqrt {ax+b}}+bx^{n}{\sqrt {ax+b}}-nb\int x^{n-1}{\sqrt {ax+b}}\,dx\right)} Remove adsLoading related searches...Wikiwand - on Seamless Wikipedia browsing. On steroids.Remove ads
, за случаите 
, виж следващата секция:
, трябва да се вземе само положителната стойност на 
.
![{\displaystyle \int {\frac {dx}{s^{5}}}={\frac {1}{a^{4}}}\left[{\frac {x}{s}}-{\frac {1}{3}}{\frac {x^{3}}{s^{3}}}\right]}](//wikimedia.org/api/rest_v1/media/math/render/svg/054a5959ce5e03cf279c1b29dff2ba014ac6dcde)
![{\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]}](//wikimedia.org/api/rest_v1/media/math/render/svg/86843311de7fc72bc01f87742445f7c4b88899e9)
![{\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]}](//wikimedia.org/api/rest_v1/media/math/render/svg/ca32b3a8d7f9040840f5d1de3467129edff0d80b)
![{\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]}](//wikimedia.org/api/rest_v1/media/math/render/svg/96ea4b7b2973dd3e2affa09931a8bf41316161f1)
![{\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]}](//wikimedia.org/api/rest_v1/media/math/render/svg/239ff6c41a3342440c712b9f0c4940e8e6a000d2)
![{\displaystyle \int {\frac {dx}{s^{5}}}={\frac {1}{a^{4}}}\left[{\frac {x}{s}}-{\frac {1}{3}}{\frac {x^{3}}{s^{3}}}\right]}](http://wikimedia.org/api/rest_v1/media/math/render/svg/054a5959ce5e03cf279c1b29dff2ba014ac6dcde)
![{\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]}](http://wikimedia.org/api/rest_v1/media/math/render/svg/86843311de7fc72bc01f87742445f7c4b88899e9)
![{\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]}](http://wikimedia.org/api/rest_v1/media/math/render/svg/ca32b3a8d7f9040840f5d1de3467129edff0d80b)
![{\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]}](http://wikimedia.org/api/rest_v1/media/math/render/svg/96ea4b7b2973dd3e2affa09931a8bf41316161f1)
![{\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]}](http://wikimedia.org/api/rest_v1/media/math/render/svg/239ff6c41a3342440c712b9f0c4940e8e6a000d2)
