Daftar integral dari fungsi rasional

tabel integral.

∫ ( a x + b ) n d x {\displaystyle \int (ax+b)^{n}dx} = ( a x + b ) n + 1 a ( n + 1 ) (untuk  n ≠ − 1 ) {\displaystyle ={\frac {(ax+b)^{n+1}}{a(n+1)}}\qquad {\mbox{(untuk }}n\neq -1{\mbox{)}}\,\!}
∫ 1 a x + b d x {\displaystyle \int {\frac {1}{ax+b}}dx} = 1 a ln ⁡ | a x + b | {\displaystyle ={\frac {1}{a}}\ln \left|ax+b\right|}
∫ x ( a x + b ) n d x {\displaystyle \int x(ax+b)^{n}dx} = a ( n + 1 ) x − b a 2 ( n + 1 ) ( n + 2 ) ( a x + b ) n + 1 (untuk  n ∉ { − 1 , − 2 } ) {\displaystyle ={\frac {a(n+1)x-b}{a^{2}(n+1)(n+2)}}(ax+b)^{n+1}\qquad {\mbox{(untuk }}n\not \in \{-1,-2\}{\mbox{)}}}
∫ x a x + b d x {\displaystyle \int {\frac {x}{ax+b}}dx} = x a − b a 2 ln ⁡ | a x + b | {\displaystyle ={\frac {x}{a}}-{\frac {b}{a^{2}}}\ln \left|ax+b\right|}
∫ x ( a x + b ) 2 d x {\displaystyle \int {\frac {x}{(ax+b)^{2}}}dx} = b a 2 ( a x + b ) + 1 a 2 ln ⁡ | a x + b | {\displaystyle ={\frac {b}{a^{2}(ax+b)}}+{\frac {1}{a^{2}}}\ln \left|ax+b\right|}
∫ x ( a x + b ) n d x {\displaystyle \int {\frac {x}{(ax+b)^{n}}}dx} = a ( 1 − n ) x − b a 2 ( n − 1 ) ( n − 2 ) ( a x + b ) n − 1 (untuk  n ∉ { 1 , 2 } ) {\displaystyle ={\frac {a(1-n)x-b}{a^{2}(n-1)(n-2)(ax+b)^{n-1}}}\qquad {\mbox{(untuk }}n\not \in \{1,2\}{\mbox{)}}}
∫ x 2 a x + b d x {\displaystyle \int {\frac {x^{2}}{ax+b}}dx} = 1 a 3 ( ( a x + b ) 2 2 − 2 b ( a x + b ) + b 2 ln ⁡ | a x + b | ) {\displaystyle ={\frac {1}{a^{3}}}\left({\frac {(ax+b)^{2}}{2}}-2b(ax+b)+b^{2}\ln \left|ax+b\right|\right)}
∫ x 2 ( a x + b ) 2 d x {\displaystyle \int {\frac {x^{2}}{(ax+b)^{2}}}dx} = 1 a 3 ( a x + b − 2 b ln ⁡ | a x + b | − b 2 a x + b ) {\displaystyle ={\frac {1}{a^{3}}}\left(ax+b-2b\ln \left|ax+b\right|-{\frac {b^{2}}{ax+b}}\right)}
∫ x 2 ( a x + b ) 3 d x {\displaystyle \int {\frac {x^{2}}{(ax+b)^{3}}}dx} = 1 a 3 ( ln ⁡ | a x + b | + 2 b a x + b − b 2 2 ( a x + b ) 2 ) {\displaystyle ={\frac {1}{a^{3}}}\left(\ln \left|ax+b\right|+{\frac {2b}{ax+b}}-{\frac {b^{2}}{2(ax+b)^{2}}}\right)}
∫ x 2 ( a x + b ) n d x {\displaystyle \int {\frac {x^{2}}{(ax+b)^{n}}}dx} = 1 a 3 ( − ( a x + b ) 3 − n ( n − 3 ) + 2 b ( a + b ) 2 − n ( n − 2 ) − b 2 ( a x + b ) 1 − n ( n − 1 ) ) (untuk  n ∉ { 1 , 2 , 3 } ) {\displaystyle ={\frac {1}{a^{3}}}\left(-{\frac {(ax+b)^{3-n}}{(n-3)}}+{\frac {2b(a+b)^{2-n}}{(n-2)}}-{\frac {b^{2}(ax+b)^{1-n}}{(n-1)}}\right)\qquad {\mbox{(untuk }}n\not \in \{1,2,3\}{\mbox{)}}}
∫ 1 x ( a x + b ) d x {\displaystyle \int {\frac {1}{x(ax+b)}}dx} = − 1 b ln ⁡ | a x + b x | {\displaystyle =-{\frac {1}{b}}\ln \left|{\frac {ax+b}{x}}\right|}
∫ 1 x 2 ( a x + b ) d x {\displaystyle \int {\frac {1}{x^{2}(ax+b)}}dx} = − 1 b x + a b 2 ln ⁡ | a x + b x | {\displaystyle =-{\frac {1}{bx}}+{\frac {a}{b^{2}}}\ln \left|{\frac {ax+b}{x}}\right|}
∫ 1 x 2 ( a x + b ) 2 d x {\displaystyle \int {\frac {1}{x^{2}(ax+b)^{2}}}dx} = − a ( 1 b 2 ( a x + b ) + 1 a b 2 x − 2 b 3 ln ⁡ | a x + b x | ) {\displaystyle =-a\left({\frac {1}{b^{2}(ax+b)}}+{\frac {1}{ab^{2}x}}-{\frac {2}{b^{3}}}\ln \left|{\frac {ax+b}{x}}\right|\right)}
∫ 1 x 2 + a 2 d x {\displaystyle \int {\frac {1}{x^{2}+a^{2}}}dx} = 1 a arctan ⁡ x a {\displaystyle ={\frac {1}{a}}\arctan {\frac {x}{a}}\,\!}
∫ 1 x 2 − a 2 d x = {\displaystyle \int {\frac {1}{x^{2}-a^{2}}}dx=}
  • − 1 a a r c t a n h x a = 1 2 a ln ⁡ a − x a + x (untuk  | x | < | a | ) {\displaystyle -{\frac {1}{a}}\,\mathrm {arctanh} {\frac {x}{a}}={\frac {1}{2a}}\ln {\frac {a-x}{a+x}}\qquad {\mbox{(untuk }}|x|<|a|{\mbox{)}}\,\!}
  • − 1 a a r c c o t h x a = 1 2 a ln ⁡ x − a x + a (untuk  | x | > | a | ) {\displaystyle -{\frac {1}{a}}\,\mathrm {arccoth} {\frac {x}{a}}={\frac {1}{2a}}\ln {\frac {x-a}{x+a}}\qquad {\mbox{(untuk }}|x|>|a|{\mbox{)}}\,\!}
∫ 1 a x 2 + b x + c d x = {\displaystyle \int {\frac {1}{ax^{2}+bx+c}}dx=}
  • 2 4 a c − b 2 arctan ⁡ 2 a x + b 4 a c − b 2 (untuk  4 a c − b 2 > 0 ) {\displaystyle {\frac {2}{\sqrt {4ac-b^{2}}}}\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}\qquad {\mbox{(untuk }}4ac-b^{2}>0{\mbox{)}}}
  • − 2 b 2 − 4 a c a r c t a n h 2 a x + b b 2 − 4 a c = 1 b 2 − 4 a c ln ⁡ | 2 a x + b − b 2 − 4 a c 2 a x + b + b 2 − 4 a c | (untuk  4 a c − b 2 < 0 ) {\displaystyle -{\frac {2}{\sqrt {b^{2}-4ac}}}\,\mathrm {arctanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}={\frac {1}{\sqrt {b^{2}-4ac}}}\ln \left|{\frac {2ax+b-{\sqrt {b^{2}-4ac}}}{2ax+b+{\sqrt {b^{2}-4ac}}}}\right|\qquad {\mbox{(untuk }}4ac-b^{2}<0{\mbox{)}}}
  • − 2 2 a x + b (untuk  4 a c − b 2 = 0 ) {\displaystyle -{\frac {2}{2ax+b}}\qquad {\mbox{(untuk }}4ac-b^{2}=0{\mbox{)}}}
∫ x a x 2 + b x + c d x {\displaystyle \int {\frac {x}{ax^{2}+bx+c}}dx} = 1 2 a ln ⁡ | a x 2 + b x + c | − b 2 a ∫ d x a x 2 + b x + c {\displaystyle ={\frac {1}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {b}{2a}}\int {\frac {dx}{ax^{2}+bx+c}}}
∫ m x + n a x 2 + b x + c d x = {\displaystyle \int {\frac {mx+n}{ax^{2}+bx+c}}dx=}
  • m 2 a ln ⁡ | a x 2 + b x + c | + 2 a n − b m a 4 a c − b 2 arctan ⁡ 2 a x + b 4 a c − b 2 (untuk  4 a c − b 2 > 0 ) {\displaystyle {\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|+{\frac {2an-bm}{a{\sqrt {4ac-b^{2}}}}}\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}\qquad {\mbox{(untuk }}4ac-b^{2}>0{\mbox{)}}}
  • m 2 a ln ⁡ | a x 2 + b x + c | + 2 a n − b m a b 2 − 4 a c a r t a n h 2 a x + b b 2 − 4 a c (untuk  4 a c − b 2 < 0 ) {\displaystyle {\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|+{\frac {2an-bm}{a{\sqrt {b^{2}-4ac}}}}\,\mathrm {artanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}\qquad {\mbox{(untuk }}4ac-b^{2}<0{\mbox{)}}}
  • m 2 a ln ⁡ | a x 2 + b x + c | − 2 a n − b m a ( 2 a x + b ) (untuk  4 a c − b 2 = 0 ) {\displaystyle {\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {2an-bm}{a(2ax+b)}}\,\,\,\,\,\,\,\,\,\,\qquad {\mbox{(untuk }}4ac-b^{2}=0{\mbox{)}}}
∫ 1 ( a x 2 + b x + c ) n d x = 2 a x + b ( n − 1 ) ( 4 a c − b 2 ) ( a x 2 + b x + c ) n − 1 + ( 2 n − 3 ) 2 a ( n − 1 ) ( 4 a c − b 2 ) ∫ 1 ( a x 2 + b x + c ) n − 1 d x {\displaystyle \int {\frac {1}{(ax^{2}+bx+c)^{n}}}dx={\frac {2ax+b}{(n-1)(4ac-b^{2})(ax^{2}+bx+c)^{n-1}}}+{\frac {(2n-3)2a}{(n-1)(4ac-b^{2})}}\int {\frac {1}{(ax^{2}+bx+c)^{n-1}}}dx\,\!}
∫ x ( a x 2 + b x + c ) n d x = b x + 2 c ( n − 1 ) ( 4 a c − b 2 ) ( a x 2 + b x + c ) n − 1 − b ( 2 n − 3 ) ( n − 1 ) ( 4 a c − b 2 ) ∫ 1 ( a x 2 + b x + c ) n − 1 d x {\displaystyle \int {\frac {x}{(ax^{2}+bx+c)^{n}}}dx={\frac {bx+2c}{(n-1)(4ac-b^{2})(ax^{2}+bx+c)^{n-1}}}-{\frac {b(2n-3)}{(n-1)(4ac-b^{2})}}\int {\frac {1}{(ax^{2}+bx+c)^{n-1}}}dx\,\!}
∫ 1 x ( a x 2 + b x + c ) d x = 1 2 c ln ⁡ | x 2 a x 2 + b x + c | − b 2 c ∫ 1 a x 2 + b x + c d x {\displaystyle \int {\frac {1}{x(ax^{2}+bx+c)}}dx={\frac {1}{2c}}\ln \left|{\frac {x^{2}}{ax^{2}+bx+c}}\right|-{\frac {b}{2c}}\int {\frac {1}{ax^{2}+bx+c}}dx}

Fungsi rasional apapun dapat diintegrasikan melalui persamaan-persamaan di atas dengan memanfaatkan integrasi parsial, dengan menguraikan fungsi rasional menjadi penjumlahan fungsi-fungsi dalam bentuk

e x + f ( a x 2 + b x + c ) n {\displaystyle {\frac {ex+f}{\left(ax^{2}+bx+c\right)^{n}}}} .

Bacaan lebih lanjut

  • Kurnianingsih, Sri (2007). Matematika SMA dan MA 3A Untuk Kelas XII Semester 1 Program IPA. Jakarta: Esis/Erlangga. ISBN 979-734-504-1.  Parameter |coauthors= yang tidak diketahui mengabaikan (|author= yang disarankan) (bantuan) (Indonesia)
  • Kurnianingsih, Sri (2007). Matematika SMA dan MA 3A Untuk Kelas XII Semester 1 Program IPS. Jakarta: Esis/Erlangga. ISBN 979-734-567-X.  Parameter |coauthors= yang tidak diketahui mengabaikan (|author= yang disarankan) (bantuan) (Indonesia)
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Daftar integral
Fungsi rasional • Fungsi irrasional • Fungsi trigonometri • Invers trigonometri • Fungsi hiperbolik • Invers hiperbolik • Fungsi eksponensial • Fungsi logaritmik