The following is a list of integrals (antiderivative functions) of irrational functions . For a complete list of integral functions, see lists of integrals . Throughout this article the constant of integration is omitted for brevity.
Integrals involving r = √a 2 + x 2 [ edit ] ∫ 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 + 3 8 a 2 x r + 3 8 a 4 ln ( x + r ) {\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)} ∫ 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 2 r 2 n + 3 2 n + 3 {\displaystyle \int x^{3}r^{2n+1}\,dx={\frac {r^{2n+5}}{2n+5}}-{\frac {a^{2}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 arsinh a x {\displaystyle \int {\frac {r\,dx}{x}}=r-a\ln \left|{\frac {a+r}{x}}\right|=r-a\,\operatorname {arsinh} {\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 = arsinh x a = ln ( x + r a ) {\displaystyle \int {\frac {dx}{r}}=\operatorname {arsinh} {\frac {x}{a}}=\ln \left({\frac {x+r}{a}}\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 arsinh x a = x 2 r − a 2 2 ln ( x + r a ) {\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)} ∫ d x x r = − 1 a arsinh a x = − 1 a ln | a + r x | {\displaystyle \int {\frac {dx}{xr}}=-{\frac {1}{a}}\,\operatorname {arsinh} {\frac {a}{x}}=-{\frac {1}{a}}\ln \left|{\frac {a+r}{x}}\right|} Integrals involving s = √x 2 − a 2 [ edit ] Assume x 2 > a 2 (for x 2 < a 2 , see next section):
∫ s d x = 1 2 ( x s − a 2 ln | x + s | ) {\displaystyle \int s\,dx={\frac {1}{2}}\left(xs-a^{2}\ln \left|x+s\right|\right)} ∫ x s d x = 1 3 s 3 {\displaystyle \int xs\,dx={\frac {1}{3}}s^{3}} ∫ s d x x = s − | a | arccos | a x | {\displaystyle \int {\frac {s\,dx}{x}}=s-|a|\arccos \left|{\frac {a}{x}}\right|} ∫ d x s = ln | x + s a | . {\displaystyle \int {\frac {dx}{s}}=\ln \left|{\frac {x+s}{a}}\right|\,.} ln | x + s a | = sgn ( x ) arcosh | x a | = 1 2 ln ( x + s x − s ) , {\displaystyle \ln \left|{\frac {x+s}{a}}\right|=\operatorname {sgn} (x)\,\operatorname {arcosh} \left|{\frac {x}{a}}\right|={\frac {1}{2}}\ln \left({\frac {x+s}{x-s}}\right)\,,} arcosh | x a | {\displaystyle \operatorname {arcosh} \left|{\frac {x}{a}}\right|} ∫ d x x s = 1 a arcsec | x a | {\displaystyle \int {\frac {dx}{xs}}={\frac {1}{a}}\operatorname {arcsec} \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]} Integrals involving u = √a 2 − x 2 [ edit ] ∫ u d x = 1 2 ( x u + a 2 arcsin x a ) ( | x | ≤ | a | ) {\displaystyle \int u\,dx={\frac {1}{2}}\left(xu+a^{2}\arcsin {\frac {x}{a}}\right)\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}} ∫ x u d x = − 1 3 u 3 ( | x | ≤ | a | ) {\displaystyle \int xu\,dx=-{\frac {1}{3}}u^{3}\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}} ∫ 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\,dx=-{\frac {x}{4}}u^{3}+{\frac {a^{2}}{8}}(xu+a^{2}\arcsin {\frac {x}{a}})\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}} ∫ u d x x = u − a ln | a + u x | ( | x | ≤ | a | ) {\displaystyle \int {\frac {u\,dx}{x}}=u-a\ln \left|{\frac {a+u}{x}}\right|\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}} ∫ d x u = arcsin x a ( | x | ≤ | a | ) {\displaystyle \int {\frac {dx}{u}}=\arcsin {\frac {x}{a}}\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}} ∫ x 2 d x u = 1 2 ( − x u + a 2 arcsin x a ) ( | x | ≤ | a | ) {\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{)}}} ∫ u d x = 1 2 ( x u − sgn x arcosh | x a | ) (for | x | ≥ | a | ) {\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{)}}} ∫ x u d x = − u ( | x | ≤ | a | ) {\displaystyle \int {\frac {x}{u}}\,dx=-u\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}} Integrals involving R = √ax 2 + bx + c [ edit ] Assume (ax 2 + bx + c ) cannot be reduced to the following expression (px + q )2 for some p and q .
∫ 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 arsinh 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}}}\,\operatorname {arsinh} {\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 | , c > 0 {\displaystyle \int {\frac {dx}{xR}}=-{\frac {1}{\sqrt {c}}}\ln \left|{\frac {2{\sqrt {c}}R+bx+2c}{x}}\right|,~c>0} ∫ d x x R = − 1 c arsinh ( b x + 2 c | x | 4 a c − b 2 ) , c < 0 {\displaystyle \int {\frac {dx}{xR}}=-{\frac {1}{\sqrt {c}}}\operatorname {arsinh} \left({\frac {bx+2c}{|x|{\sqrt {4ac-b^{2}}}}}\right),~c<0} ∫ d x x R = 1 − c arcsin ( b x + 2 c | x | b 2 − 4 a c ) , c < 0 , b 2 − 4 a c > 0 {\displaystyle \int {\frac {dx}{xR}}={\frac {1}{\sqrt {-c}}}\operatorname {arcsin} \left({\frac {bx+2c}{|x|{\sqrt {b^{2}-4ac}}}}\right),~c<0,b^{2}-4ac>0} ∫ d x x R = − 2 b x ( a x 2 + b x ) , c = 0 {\displaystyle \int {\frac {dx}{xR}}=-{\frac {2}{bx}}\left({\sqrt {ax^{2}+bx}}\right),~c=0} ∫ x 2 R d x = 2 a x − 3 b 4 a 2 R + 3 b 2 − 4 a c 8 a 2 ∫ d x R {\displaystyle \int {\frac {x^{2}}{R}}\,dx={\frac {2ax-3b}{4a^{2}}}R+{\frac {3b^{2}-4ac}{8a^{2}}}\int {\frac {dx}{R}}} ∫ d x x 2 R = − R c x − b 2 c ∫ d x x R {\displaystyle \int {\frac {dx}{x^{2}R}}=-{\frac {R}{cx}}-{\frac {b}{2c}}\int {\frac {dx}{xR}}} ∫ R d x = 2 a x + b 4 a R + 4 a c − b 2 8 a ∫ d x R {\displaystyle \int R\,dx={\frac {2ax+b}{4a}}R+{\frac {4ac-b^{2}}{8a}}\int {\frac {dx}{R}}} ∫ x R d x = R 3 3 a − b ( 2 a x + b ) 8 a 2 R − b ( 4 a c − b 2 ) 16 a 2 ∫ d x R {\displaystyle \int xR\,dx={\frac {R^{3}}{3a}}-{\frac {b(2ax+b)}{8a^{2}}}R-{\frac {b(4ac-b^{2})}{16a^{2}}}\int {\frac {dx}{R}}} ∫ x 2 R d x = 6 a x − 5 b 24 a 2 R 3 + 5 b 2 − 4 a c 16 a 2 ∫ R d x {\displaystyle \int x^{2}R\,dx={\frac {6ax-5b}{24a^{2}}}R^{3}+{\frac {5b^{2}-4ac}{16a^{2}}}\int R\,dx} ∫ R x d x = R + b 2 ∫ d x R + c ∫ d x x R {\displaystyle \int {\frac {R}{x}}\,dx=R+{\frac {b}{2}}\int {\frac {dx}{R}}+c\int {\frac {dx}{xR}}} ∫ R x 2 d x = − R x + a ∫ d x R + b 2 ∫ d x x R {\displaystyle \int {\frac {R}{x^{2}}}\,dx=-{\frac {R}{x}}+a\int {\frac {dx}{R}}+{\frac {b}{2}}\int {\frac {dx}{xR}}} ∫ x 2 d x R 3 = ( 2 b 2 − 4 a c ) x + 2 b c a ( 4 a c − b 2 ) R + 1 a ∫ d x R {\displaystyle \int {\frac {x^{2}\,dx}{R^{3}}}={\frac {(2b^{2}-4ac)x+2bc}{a(4ac-b^{2})R}}+{\frac {1}{a}}\int {\frac {dx}{R}}} Integrals involving S = √ax + b [ edit ] ∫ S d x = 2 S 3 3 a {\displaystyle \int S\,dx={\frac {2S^{3}}{3a}}} ∫ d x S = 2 S a {\displaystyle \int {\frac {dx}{S}}={\frac {2S}{a}}} ∫ d x x S = { − 2 b arcoth ( S b ) (for b > 0 , a x > 0 ) − 2 b artanh ( S b ) (for b > 0 , a x < 0 ) 2 − b arctan ( S − b ) (for b < 0 ) {\displaystyle \int {\frac {dx}{xS}}={\begin{cases}-{\dfrac {2}{\sqrt {b}}}\operatorname {arcoth} \left({\dfrac {S}{\sqrt {b}}}\right)&{\mbox{(for }}b>0,\quad ax>0{\mbox{)}}\\-{\dfrac {2}{\sqrt {b}}}\operatorname {artanh} \left({\dfrac {S}{\sqrt {b}}}\right)&{\mbox{(for }}b>0,\quad ax<0{\mbox{)}}\\{\dfrac {2}{\sqrt {-b}}}\arctan \left({\dfrac {S}{\sqrt {-b}}}\right)&{\mbox{(for }}b<0{\mbox{)}}\\\end{cases}}} ∫ S x d x = { 2 ( S − b arcoth ( S b ) ) (for b > 0 , a x > 0 ) 2 ( S − b artanh ( S b ) ) (for b > 0 , a x < 0 ) 2 ( S − − b arctan ( S − b ) ) (for b < 0 ) {\displaystyle \int {\frac {S}{x}}\,dx={\begin{cases}2\left(S-{\sqrt {b}}\,\operatorname {arcoth} \left({\dfrac {S}{\sqrt {b}}}\right)\right)&{\mbox{(for }}b>0,\quad ax>0{\mbox{)}}\\2\left(S-{\sqrt {b}}\,\operatorname {artanh} \left({\dfrac {S}{\sqrt {b}}}\right)\right)&{\mbox{(for }}b>0,\quad ax<0{\mbox{)}}\\2\left(S-{\sqrt {-b}}\arctan \left({\dfrac {S}{\sqrt {-b}}}\right)\right)&{\mbox{(for }}b<0{\mbox{)}}\\\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}}\,dx={\frac {2}{a(2n+1)}}\left(x^{n}S-bn\int {\frac {x^{n-1}}{S}}\,dx\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\,dx={\frac {2}{a(2n+3)}}\left(x^{n}S^{3}-nb\int x^{n-1}S\,dx\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}}\,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)} References [ edit ] Abramowitz, Milton; Stegun, Irene A., eds. (1972). "Chapter 3" . Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables Gradshteyn, Izrail Solomonovich ; Ryzhik, Iosif Moiseevich ; Geronimus, Yuri Veniaminovich ; Tseytlin, Michail Yulyevich ; Jeffrey, Alan (2015) [October 2014]. Zwillinger, Daniel; Moll, Victor Hugo (eds.). Table of Integrals, Series, and Products Academic Press, Inc. ISBN 978-0-12-384933-5 LCCN 2014010276 .Peirce, Benjamin Osgood (1929) [1899]. "Chapter 3". A Short Table of Integrals (3rd revised ed.). Boston: Ginn and Co. pp. 16–30.
![{\displaystyle \int {\frac {dx}{s^{5}}}={\frac {1}{a^{4}}}\left[{\frac {x}{s}}-{\frac {1}{3}}{\frac {x^{3}}{s^{3}}}\right]}](https://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]}](https://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]}](https://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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6a98057cf3f3d6b7025114445c972bb6b7b7af9d)
![{\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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9cce4b87e7a47ce42042803038139f830afd5d37)
