Dit artikel bevat een lijst van integralen van irrationale functies. Integralen zijn het onderwerp van studie van de integraalrekening. De integralen in de lijst hieronder zijn veel voorkomende integralen van een functie onder de wortel. Er wordt van alle integralen de primitieve functie gegeven, maar de integratieconstante is in de uitkomst steeds weggelaten. Integralen met 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\ \mathrm {d} x={\tfrac {1}{2}}(xr+a^{2}\ \ln(x+r))} ∫ r 3 d x = 1 4 x r 3 + 3 8 a 2 x r + 3 8 a 4 ln ( x + r ) {\displaystyle \int r^{3}\ \mathrm {d} x={\tfrac {1}{4}}xr^{3}+{\tfrac {3}{8}}a^{2}xr+{\tfrac {3}{8}}a^{4}\ln(x+r)} ∫ 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}\ \mathrm {d} x={\tfrac {1}{6}}xr^{5}+{\tfrac {5}{24}}a^{2}xr^{3}+{\tfrac {5}{16}}a^{4}xr+{\tfrac {5}{16}}a^{6}\ln(x+r)} ∫ x r d x = 1 3 r 3 {\displaystyle \int xr\ \mathrm {d} x={\tfrac {1}{3}}r^{3}} ∫ x r 3 d x = 1 5 r 5 {\displaystyle \int xr^{3}\ \mathrm {d} x={\tfrac {1}{5}}r^{5}} ∫ x r 2 n + 1 d x = r 2 n + 3 2 n + 3 {\displaystyle \int xr^{2n+1}\ \mathrm {d} x={\frac {r^{2n+3}}{2n+3}}} ∫ x 2 r d x = 1 4 x r 3 − 1 8 a 2 x r − 1 8 a 4 ln ( x + r ) {\displaystyle \int x^{2}r\ \mathrm {d} x={\tfrac {1}{4}}xr^{3}-{\tfrac {1}{8}}a^{2}xr-{\tfrac {1}{8}}a^{4}\ln(x+r)} ∫ x 2 r 3 d x = 1 6 x r 5 − 1 24 a 2 x r 3 − 1 16 a 4 x r − 1 16 a 6 ln ( x + r ) {\displaystyle \int x^{2}r^{3}\ \mathrm {d} x={\tfrac {1}{6}}xr^{5}-{\tfrac {1}{24}}a^{2}xr^{3}-{\tfrac {1}{16}}a^{4}xr-{\tfrac {1}{16}}a^{6}\ln(x+r)} ∫ x 3 r d x = 1 5 r 5 − 1 3 a 2 r 3 {\displaystyle \int x^{3}r\ \mathrm {d} x={\tfrac {1}{5}}r^{5}-{\tfrac {1}{3}}a^{2}r^{3}} ∫ x 3 r 3 d x = 1 7 r 7 − 1 5 a 2 r 5 {\displaystyle \int x^{3}r^{3}\ \mathrm {d} x={\tfrac {1}{7}}r^{7}-{\tfrac {1}{5}}a^{2}r^{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}\ \mathrm {d} x={\frac {r^{2n+5}}{2n+5}}-{\frac {a^{3}r^{2n+3}}{2n+3}}} ∫ x 4 r d x = 1 6 x 3 r 3 − 1 8 a 2 x r 3 + 1 16 a 4 x r + 1 16 a 6 ln ( x + r ) {\displaystyle \int x^{4}r\ \mathrm {d} x={\tfrac {1}{6}}x^{3}r^{3}-{\tfrac {1}{8}}a^{2}xr^{3}+{\tfrac {1}{16}}a^{4}xr+{\tfrac {1}{16}}a^{6}\ln(x+r)} ∫ x 4 r 3 d x = 1 8 x 3 r 5 − 1 16 a 2 x r 5 + 1 64 a 4 x r 3 + 3 128 a 6 x r + 3 128 a 8 ln ( x + r ) {\displaystyle \int x^{4}r^{3}\ \mathrm {d} x={\tfrac {1}{8}}x^{3}r^{5}-{\tfrac {1}{16}}a^{2}xr^{5}+{\tfrac {1}{64}}a^{4}xr^{3}+{\tfrac {3}{128}}a^{6}xr+{\tfrac {3}{128}}a^{8}\ln(x+r)} ∫ x 5 r d x = 1 7 r 7 − 2 5 a 2 r 5 + 1 3 a 4 r 3 {\displaystyle \int x^{5}r\ \mathrm {d} x={\tfrac {1}{7}}r^{7}-{\tfrac {2}{5}}a^{2}r^{5}+{\tfrac {1}{3}}a^{4}r^{3}} ∫ x 5 r 3 d x = 1 9 r 9 − 2 7 a 2 r 7 + 1 5 a 4 r 5 {\displaystyle \int x^{5}r^{3}\ \mathrm {d} x={\tfrac {1}{9}}r^{9}-{\tfrac {2}{7}}a^{2}r^{7}+{\tfrac {1}{5}}a^{4}r^{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}\ \mathrm {d} x={\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 arsinh a x {\displaystyle \int {\frac {r\ \mathrm {d} x}{x}}=r-a\ln \left|{\frac {a+r}{x}}\right|=r-a\ \operatorname {arsinh} {\frac {a}{x}}} ∫ r 3 d x x = 1 3 r 3 + a 2 r − a 3 ln | a + r x | {\displaystyle \int {\frac {r^{3}\ \mathrm {d} x}{x}}={\tfrac {1}{3}}r^{3}+a^{2}r-a^{3}\ln \left|{\frac {a+r}{x}}\right|} ∫ r 5 d x x = 1 5 r 5 + 1 3 a 2 r 3 + a 4 r − a 5 ln | a + r x | {\displaystyle \int {\frac {r^{5}\ \mathrm {d} x}{x}}={\tfrac {1}{5}}r^{5}+{\tfrac {1}{3}}a^{2}r^{3}+a^{4}r-a^{5}\ln \left|{\frac {a+r}{x}}\right|} ∫ r 7 d x x = 1 7 r 7 + 1 5 a 2 r 5 + 1 3 a 4 r 3 + a 6 r − a 7 ln | a + r x | {\displaystyle \int {\frac {r^{7}\ \mathrm {d} x}{x}}={\tfrac {1}{7}}r^{7}+{\tfrac {1}{5}}a^{2}r^{5}+{\tfrac {1}{3}}a^{4}r^{3}+a^{6}r-a^{7}\ln \left|{\frac {a+r}{x}}\right|} ∫ d x r = arsinh x a = ln ( x + r a ) {\displaystyle \int {\frac {\mathrm {d} x}{r}}=\operatorname {arsinh} {\frac {x}{a}}=\ln \left({\frac {x+r}{a}}\right)} ∫ d x r 3 = x a 2 r {\displaystyle \int {\frac {\mathrm {d} x}{r^{3}}}={\frac {x}{a^{2}r}}} ∫ x d x r = r {\displaystyle \int {\frac {x\ \ \mathrm {d} x}{r}}=r} ∫ x d x r 3 = − 1 r {\displaystyle \int {\frac {x\ \ \mathrm {d} x}{r^{3}}}=-{\frac {1}{r}}} ∫ x 2 d x r = 1 2 x r − 1 2 a 2 arsinh x a = 1 2 x r − 1 2 a 2 ln ( x + r a ) {\displaystyle \int {\frac {x^{2}\ \mathrm {d} x}{r}}={\tfrac {1}{2}}xr-{\tfrac {1}{2}}a^{2}\ \operatorname {arsinh} {\frac {x}{a}}={\tfrac {1}{2}}xr-{\tfrac {1}{2}}a^{2}\ln \left({\frac {x+r}{a}}\right)} ∫ d x x r = − 1 a arsinh a x = − 1 a ln | a + r x | {\displaystyle \int {\frac {\mathrm {d} x}{xr}}=-{\frac {1}{a}}\ \operatorname {arsinh} {\frac {a}{x}}=-{\frac {1}{a}}\ln \left|{\frac {a+r}{x}}\right|} Integralen met s = x 2 − a 2 {\displaystyle s={\sqrt {x^{2}-a^{2}}}} Voor de volgende integralen is x 2 > a 2 {\displaystyle x^{2}>a^{2}} . ∫ s d x = 1 2 ( x s − a 2 ln ( x + s ) ) {\displaystyle \int s\ \mathrm {d} x={\tfrac {1}{2}}(xs-a^{2}\ln(x+s))} ∫ x s d x = 1 3 s 3 {\displaystyle \int xs\ \mathrm {d} x={\tfrac {1}{3}}s^{3}} ∫ s d x x = s − a arccos | a x | {\displaystyle \int {\frac {s\ \mathrm {d} x}{x}}=s-a\arccos \left|{\frac {a}{x}}\right|} ∫ d x s = ∫ d x x 2 − a 2 = ln | x + s a | {\displaystyle \int {\frac {\mathrm {d} x}{s}}=\int {\frac {\mathrm {d} x}{\sqrt {x^{2}-a^{2}}}}=\ln \left|{\frac {x+s}{a}}\right|} Houd hier rekening met het feit dat ln | x + s a | = s g n ( x ) arcosh | x a | = 1 2 ln ( x + s x − s ) {\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)} , waarbij alleen naar de positieve waarde moet worden gekeken, namelijk arcosh | x a | {\displaystyle \operatorname {arcosh} \left|{\frac {x}{a}}\right|} ∫ x d x s = s {\displaystyle \int {\frac {x\ \mathrm {d} x}{s}}=s} ∫ x d x s 3 = − 1 s {\displaystyle \int {\frac {x\ \mathrm {d} x}{s^{3}}}=-{\frac {1}{s}}} ∫ x d x s 5 = − 1 3 s 3 {\displaystyle \int {\frac {x\ \mathrm {d} x}{s^{5}}}=-{\frac {1}{3s^{3}}}} ∫ x d x s 7 = − 1 5 s 5 {\displaystyle \int {\frac {x\ \mathrm {d} x}{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\ \mathrm {d} x}{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}\ \mathrm {d} x}{s^{2n+1}}}=-{\frac {1}{2n-1}}{\frac {x^{2m-1}}{s^{2n-1}}}+{\frac {2m-1}{2n-1}}\int {\frac {x^{2m-2}\ \mathrm {d} x}{s^{2n-1}}}} ∫ x 2 d x s = 1 2 x s + 1 2 a 2 ln | x + s a | {\displaystyle \int {\frac {x^{2}\ \mathrm {d} x}{s}}={\tfrac {1}{2}}xs+{\tfrac {1}{2}}a^{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}\ \mathrm {d} x}{s^{3}}}=-{\frac {x}{s}}+\ln \left|{\frac {x+s}{a}}\right|} ∫ x 4 d x s = 1 4 x 3 s + 3 8 a 2 x s + 3 8 a 4 ln | x + s a | {\displaystyle \int {\frac {x^{4}\ \mathrm {d} x}{s}}={\tfrac {1}{4}}x^{3}s+{\tfrac {3}{8}}a^{2}xs+{\tfrac {3}{8}}a^{4}\ln \left|{\frac {x+s}{a}}\right|} ∫ x 4 d x s 3 = 1 2 x s − a 2 x s + 3 2 a 2 ln | x + s a | {\displaystyle \int {\frac {x^{4}\ \mathrm {d} x}{s^{3}}}={\tfrac {1}{2}}xs-{\frac {a^{2}x}{s}}+{\tfrac {3}{2}}a^{2}\ln \left|{\frac {x+s}{a}}\right|} ∫ x 4 d x s 5 = − x s − x 3 3 s 3 + ln | x + s a | {\displaystyle \int {\frac {x^{4}\ \mathrm {d} x}{s^{5}}}=-{\frac {x}{s}}-{\frac {x^{3}}{3s^{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}\ \mathrm {d} x}{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 n>m\geq 0} ∫ d x s 3 = − 1 a 2 x s {\displaystyle \int {\frac {\mathrm {d} x}{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 {\mathrm {d} x}{s^{5}}}={\frac {1}{a^{4}}}\left[{\frac {x}{s}}-{\tfrac {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 {\mathrm {d} x}{s^{7}}}=-{\frac {1}{a^{6}}}\left[{\frac {x}{s}}-{\tfrac {2}{3}}{\frac {x^{3}}{s^{3}}}+{\tfrac {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 {\mathrm {d} x}{s^{9}}}={\frac {1}{a^{8}}}\left[{\frac {x}{s}}-{\tfrac {3}{3}}{\frac {x^{3}}{s^{3}}}+{\tfrac {3}{5}}{\frac {x^{5}}{s^{5}}}-{\tfrac {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}\ \mathrm {d} x}{s^{5}}}=-{\frac {1}{a^{2}}}{\frac {x^{3}}{3s^{3}}}} ∫ x 2 d x s 7 = 1 a 4 [ 1 3 x 3 3 s 3 − 1 5 x 5 s 5 ] {\displaystyle \int {\frac {x^{2}\ \mathrm {d} x}{s^{7}}}={\frac {1}{a^{4}}}\left[{\tfrac {1}{3}}{\frac {x^{3}}{3s^{3}}}-{\tfrac {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}\ \mathrm {d} x}{s^{9}}}=-{\frac {1}{a^{6}}}\left[{\tfrac {1}{3}}{\frac {x^{3}}{s^{3}}}-{\tfrac {2}{5}}{\frac {x^{5}}{s^{5}}}+{\tfrac {1}{7}}{\frac {x^{7}}{s^{7}}}\right]} Integralen waarbij u = a 2 − x 2 {\displaystyle u={\sqrt {a^{2}-x^{2}}}} ∫ u d x = 1 2 ( x u + a 2 arcsin x a ) | x | ≤ | a | {\displaystyle \int u\ \mathrm {d} x={\frac {1}{2}}\left(xu+a^{2}\arcsin {\frac {x}{a}}\right)\qquad |x|\leq |a|} ∫ x u d x = − 1 3 u 3 ( | x | ≤ | a | ) {\displaystyle \int xu\ \mathrm {d} x=-{\frac {1}{3}}u^{3}\qquad (|x|\leq |a|)} ∫ x 2 u d x = − x 4 u 3 + a 2 8 ( x u + a 2 arcsin x a ) | x | ≤ | a | {\displaystyle \int x^{2}u\ \mathrm {d} x=-{\frac {x}{4}}u^{3}+{\frac {a^{2}}{8}}(xu+a^{2}\arcsin {\frac {x}{a}})\qquad |x|\leq |a|} ∫ u d x x = u − a ln | a + u x | | x | ≤ | a | {\displaystyle \int {\frac {u\ \mathrm {d} x}{x}}=u-a\ln \left|{\frac {a+u}{x}}\right|\qquad |x|\leq |a|} ∫ d x u = arcsin x a ( | x | ≤ | a | ) {\displaystyle \int {\frac {\mathrm {d} x}{u}}=\arcsin {\frac {x}{a}}\qquad (|x|\leq |a|)} ∫ x 2 d x u = 1 2 ( − x u + a 2 arcsin x a ) | x | ≤ | a | {\displaystyle \int {\frac {x^{2}\ \mathrm {d} x}{u}}={\frac {1}{2}}\left(-xu+a^{2}\arcsin {\frac {x}{a}}\right)\qquad |x|\leq |a|} ∫ u d x = 1 2 ( x u − sgn x arcosh | x a | ) voor | x | ≥ | a | {\displaystyle \int u\ \mathrm {d} x={\frac {1}{2}}\left(xu-\operatorname {sgn} x\ \operatorname {arcosh} \left|{\frac {x}{a}}\right|\right)\qquad {\mbox{voor }}|x|\geq |a|} ∫ x u d x = − u | x | ≤ | a | {\displaystyle \int {\frac {x}{u}}\ \mathrm {d} x=-u\qquad |x|\leq |a|} Integralen waarbij R = a x 2 + b x + c {\displaystyle R={\sqrt {ax^{2}+bx+c}}} Neem aan dat a x 2 + b x + c {\displaystyle ax^{2}+bx+c} niet kan worden geschreven als de verkorte vorm ( p x + q ) 2 {\displaystyle (px+q)^{2}} ∫ d x R = 1 a ln | 2 a R + 2 a x + b | voor a > 0 {\displaystyle \int {\frac {\mathrm {d} x}{R}}={\frac {1}{\sqrt {a}}}\ln \left|2{\sqrt {a}}R+2ax+b\right|\qquad {\mbox{voor }}a>0} ∫ d x R = 1 a arsinh 2 a x + b 4 a c − b 2 voor a > 0 , 4 a c − b 2 > 0 {\displaystyle \int {\frac {\mathrm {d} x}{R}}={\frac {1}{\sqrt {a}}}\ \operatorname {arsinh} {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}\qquad {\mbox{voor }}a>0,\ 4ac-b^{2}>0} ∫ d x R = 1 a ln | 2 a x + b | voor a > 0 , 4 a c − b 2 = 0 {\displaystyle \int {\frac {\mathrm {d} x}{R}}={\frac {1}{\sqrt {a}}}\ln |2ax+b|\quad {\mbox{voor }}a>0,\ 4ac-b^{2}=0} ∫ d x R = − 1 − a arcsin 2 a x + b b 2 − 4 a c voor a < 0 , 4 a c − b 2 < 0 , | 2 a x + b | < b 2 − 4 a c {\displaystyle \int {\frac {\mathrm {d} x}{R}}=-{\frac {1}{\sqrt {-a}}}\arcsin {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}\qquad {\mbox{voor }}a<0,\ 4ac-b^{2}<0,\ |2ax+b|<{\sqrt {b^{2}-4ac}}} ∫ d x R 3 = 4 a x + 2 b ( 4 a c − b 2 ) R {\displaystyle \int {\frac {\mathrm {d} x}{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 {\mathrm {d} x}{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 {\mathrm {d} x}{R^{2n+1}}}={\frac {2}{(2n-1)(4ac-b^{2})}}\left({\frac {2ax+b}{R^{2n-1}}}+4a(n-1)\int {\frac {\mathrm {d} x}{R^{2n-1}}}\right)} ∫ x R d x = R a − b 2 a ∫ d x R {\displaystyle \int {\frac {x}{R}}\ \mathrm {d} x={\frac {R}{a}}-{\frac {b}{2a}}\int {\frac {\mathrm {d} x}{R}}} ∫ x R 3 d x = − 2 b x + 4 c ( 4 a c − b 2 ) R {\displaystyle \int {\frac {x}{R^{3}}}\ \mathrm {d} x=-{\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}}}\ \mathrm {d} x=-{\frac {1}{(2n-1)aR^{2n-1}}}-{\frac {b}{2a}}\int {\frac {\mathrm {d} x}{R^{2n+1}}}} ∫ d x x R = − 1 c ln ( 2 c R + b x + 2 c x ) {\displaystyle \int {\frac {\mathrm {d} x}{xR}}=-{\frac {1}{\sqrt {c}}}\ln \left({\frac {2{\sqrt {c}}R+bx+2c}{x}}\right)} ∫ d x x R = − 1 c arsinh ( b x + 2 c | x | 4 a c − b 2 ) {\displaystyle \int {\frac {\mathrm {d} x}{xR}}=-{\frac {1}{\sqrt {c}}}\operatorname {arsinh} \left({\frac {bx+2c}{|x|{\sqrt {4ac-b^{2}}}}}\right)} Integralen waarbij S = a x + b {\displaystyle S={\sqrt {ax+b}}} ∫ S d x = 2 S 3 3 a {\displaystyle \int S{\mathrm {d} x}={\frac {2S^{3}}{3a}}} ∫ d x S = 2 S a {\displaystyle \int {\frac {\mathrm {d} x}{S}}={\frac {2S}{a}}} ∫ d x x S = { voor b > 0 , a x > 0 − 2 b a r t a n h ( S b ) voor b > 0 , a x < 0 2 − b arctan ( S − b ) voor b < 0 {\displaystyle \int {\frac {\mathrm {d} x}{xS}}={\begin{cases}{\mbox{voor }}b>0,\quad ax>0\\-{\frac {2}{\sqrt {b}}}\mathrm {artanh} \left({\frac {S}{\sqrt {b}}}\right)&{\mbox{voor }}b>0,\quad ax<0\\{\frac {2}{\sqrt {-b}}}\arctan \left({\frac {S}{\sqrt {-b}}}\right)&{\mbox{voor }}b<0\\\end{cases}}} ∫ S x d x = { 2 ( S − b a r c o t h ( S b ) ) voor b > 0 , a x > 0 2 ( S − b a r t a n h ( S b ) ) voor b > 0 , a x < 0 2 ( S − − b arctan ( S − b ) ) voor b < 0 {\displaystyle \int {\frac {S}{x}}\ \mathrm {d} x={\begin{cases}2\left(S-{\sqrt {b}}\ \mathrm {arcoth} \left({\frac {S}{\sqrt {b}}}\right)\right)&{\mbox{voor }}b>0,\quad ax>0\\2\left(S-{\sqrt {b}}\ \mathrm {artanh} \left({\frac {S}{\sqrt {b}}}\right)\right)&{\mbox{voor }}b>0,\quad ax<0\\2\left(S-{\sqrt {-b}}\arctan \left({\frac {S}{\sqrt {-b}}}\right)\right)&{\mbox{voor }}b<0\\\end{cases}}} ∫ x n S d x = 2 a ( 2 n + 1 ) ( x n S − b n ∫ x n − 1 S d x ) {\displaystyle \int {\frac {x^{n}}{S}}\ \mathrm {d} x={\frac {2}{a(2n+1)}}\left(x^{n}S-bn\int {\frac {x^{n-1}}{S}}\ \mathrm {d} x\right)} ∫ x n S d x = 2 a ( 2 n + 3 ) ( x n S 3 − n b ∫ x n − 1 S d x ) {\displaystyle \int x^{n}S\ \mathrm {d} x={\frac {2}{a(2n+3)}}\left(x^{n}S^{3}-nb\int x^{n-1}S\ \mathrm {d} x\right)} ∫ 1 x n S d x = − 1 b ( n − 1 ) ( S x n − 1 + ( n − 3 2 ) a ∫ d x x n − 1 S ) {\displaystyle \int {\frac {1}{x^{n}S}}\ \mathrm {d} x=-{\frac {1}{b(n-1)}}\left({\frac {S}{x^{n-1}}}+\left(n-{\frac {3}{2}}\right)a\int {\frac {\mathrm {d} x}{x^{n-1}S}}\right)} Literatuur S Gradshteyn, IM Ryzhik, A Jeffrey en D Zwillinger. Table of Integrals, Series, and Products, 2007. met te downloaden pdf's, ISBN 978-0-12-373637-6. Lijst van integralen rationaal · irrationaal · exponentieel · logaritmisch · goniometrisch · invers goniometrisch · hyperbolisch · invers hyperbolisch Loading related searches...Wikiwand - on Seamless Wikipedia browsing. On steroids.
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, waarbij alleen naar de positieve waarde moet worden gekeken, namelijk 
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![{\displaystyle \int {\frac {\mathrm {d} x}{s^{7}}}=-{\frac {1}{a^{6}}}\left[{\frac {x}{s}}-{\tfrac {2}{3}}{\frac {x^{3}}{s^{3}}}+{\tfrac {1}{5}}{\frac {x^{5}}{s^{5}}}\right]}](//wikimedia.org/api/rest_v1/media/math/render/svg/8a6f3edccad00497079e192ae699e960e152acb4)
![{\displaystyle \int {\frac {\mathrm {d} x}{s^{9}}}={\frac {1}{a^{8}}}\left[{\frac {x}{s}}-{\tfrac {3}{3}}{\frac {x^{3}}{s^{3}}}+{\tfrac {3}{5}}{\frac {x^{5}}{s^{5}}}-{\tfrac {1}{7}}{\frac {x^{7}}{s^{7}}}\right]}](//wikimedia.org/api/rest_v1/media/math/render/svg/e712fcf25118a8188eab07ee1cdbcf6fa4ae4659)
![{\displaystyle \int {\frac {x^{2}\ \mathrm {d} x}{s^{7}}}={\frac {1}{a^{4}}}\left[{\tfrac {1}{3}}{\frac {x^{3}}{3s^{3}}}-{\tfrac {1}{5}}{\frac {x^{5}}{s^{5}}}\right]}](//wikimedia.org/api/rest_v1/media/math/render/svg/8144c9e828d29a8e0537fd6d032a515d6ad8a19a)
![{\displaystyle \int {\frac {x^{2}\ \mathrm {d} x}{s^{9}}}=-{\frac {1}{a^{6}}}\left[{\tfrac {1}{3}}{\frac {x^{3}}{s^{3}}}-{\tfrac {2}{5}}{\frac {x^{5}}{s^{5}}}+{\tfrac {1}{7}}{\frac {x^{7}}{s^{7}}}\right]}](//wikimedia.org/api/rest_v1/media/math/render/svg/7bd7215c6045dc7365d33c5c6a50e72d33b8647b)
![{\displaystyle \int {\frac {\mathrm {d} x}{s^{5}}}={\frac {1}{a^{4}}}\left[{\frac {x}{s}}-{\tfrac {1}{3}}{\frac {x^{3}}{s^{3}}}\right]}](http://wikimedia.org/api/rest_v1/media/math/render/svg/bd705560c62f024beb029c160f94b55c41dc9422)
![{\displaystyle \int {\frac {\mathrm {d} x}{s^{7}}}=-{\frac {1}{a^{6}}}\left[{\frac {x}{s}}-{\tfrac {2}{3}}{\frac {x^{3}}{s^{3}}}+{\tfrac {1}{5}}{\frac {x^{5}}{s^{5}}}\right]}](http://wikimedia.org/api/rest_v1/media/math/render/svg/8a6f3edccad00497079e192ae699e960e152acb4)
![{\displaystyle \int {\frac {\mathrm {d} x}{s^{9}}}={\frac {1}{a^{8}}}\left[{\frac {x}{s}}-{\tfrac {3}{3}}{\frac {x^{3}}{s^{3}}}+{\tfrac {3}{5}}{\frac {x^{5}}{s^{5}}}-{\tfrac {1}{7}}{\frac {x^{7}}{s^{7}}}\right]}](http://wikimedia.org/api/rest_v1/media/math/render/svg/e712fcf25118a8188eab07ee1cdbcf6fa4ae4659)
![{\displaystyle \int {\frac {x^{2}\ \mathrm {d} x}{s^{7}}}={\frac {1}{a^{4}}}\left[{\tfrac {1}{3}}{\frac {x^{3}}{3s^{3}}}-{\tfrac {1}{5}}{\frac {x^{5}}{s^{5}}}\right]}](http://wikimedia.org/api/rest_v1/media/math/render/svg/8144c9e828d29a8e0537fd6d032a515d6ad8a19a)
![{\displaystyle \int {\frac {x^{2}\ \mathrm {d} x}{s^{9}}}=-{\frac {1}{a^{6}}}\left[{\tfrac {1}{3}}{\frac {x^{3}}{s^{3}}}-{\tfrac {2}{5}}{\frac {x^{5}}{s^{5}}}+{\tfrac {1}{7}}{\frac {x^{7}}{s^{7}}}\right]}](http://wikimedia.org/api/rest_v1/media/math/render/svg/7bd7215c6045dc7365d33c5c6a50e72d33b8647b)
