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a x + b {\displaystyle ax+b} を含む積分 ∫ 1 a x + b d x = 1 a ln ⁡ 。
a + b x {\displaystyle {\sqrt {a+bx}}} を含む積分 ∫ x a + b x d x = 2 15 b 2 ( 3 b x − 2 a ) ( a + b x ) 3 2 + C {\displaystyle \int x{\sqrt {a+bx}}\,dx={\frac {2}{15b^{2}}}(3bx-2a)(a+bx)^{\frac {3}{2}}+C} ∫ x 2 a + b x d x = 2 105 b 3 ( 15 b 2 x 2 − 12 a b x + 8 a 2 ) ( a + b x ) 3 2 + C {\displaystyle \int x^{2}{\sqrt {a+bx}}\,dx={\frac {2}{105b^{3}}}(15b^{2}x^{2}-12abx+8a^{2})(a+bx)^{\frac {3}{2}}+C} ∫ x n a + b x d x = 2 b ( 2 n + 3 ) x n ( a + b x ) 3 2 − 2 n a b ( 2 n + 3 ) ∫ x n − 1 a + b x d x {\displaystyle \int x^{n}{\sqrt {a+bx}}\,dx={\frac {2}{b(2n+3)}}x^{n}(a+bx)^{\frac {3}{2}}-{\frac {2na}{b(2n+3)}}\int x^{n-1}{\sqrt {a+bx}}dx} ∫ a + b x x d x = 2 a + b x + a ∫ 1 x a + b x d x {\displaystyle \int {\frac {\sqrt {a+bx}}{x}}\,dx=2{\sqrt {a+bx}}+a\int {\frac {1}{x{\sqrt {a+bx}}}}dx} ∫ a + b x x n d x = − 1 a ( n − 1 ) ( a + b x ) 3 2 x n − 1 − ( 2 n − 5 ) b 2 a ( n − 1 ) ∫ a + b x x n − 1 d x , n ≠ 1 {\displaystyle \int {\frac {\sqrt {a+bx}}{x^{n}}}\,dx={\frac {-1}{a(n-1)}}{\frac {(a+bx)^{\frac {3}{2}}}{x^{n-1}}}-{\frac {(2n-5)b}{2a(n-1)}}\int {\frac {\sqrt {a+bx}}{x^{n-1}}}dx,n\neq 1} ∫ 1 x a + b x d x = 1 a ln ⁡ ( a + b x − a a + b x + a ) + C , a > 0 {\displaystyle \int {\frac {1}{x{\sqrt {a+bx}}}}\,dx={\frac {1}{\sqrt {a}}}\ln \left({\frac {{\sqrt {a+bx}}-{\sqrt {a}}}{{\sqrt {a+bx}}+{\sqrt {a}}}}\right)+C,a>0} = 2 − a arctan ⁡ a + b x − a + C , a < 0 {\displaystyle ={\frac {2}{\sqrt {-a}}}\arctan {\sqrt {\frac {a+bx}{-a}}}+C,a<0} ∫ 1 x n a + b x d x = − 1 a ( n − 1 ) a + b x x n − 1 − ( 2 n − 3 ) b 2 a ( n − 1 ) ∫ 1 x n − 1 a + b x d x , n ≠ 1 {\displaystyle \int {\frac {1}{x^{n}{\sqrt {a+bx}}}}\,dx={\frac {-1}{a(n-1)}}{\frac {\sqrt {a+bx}}{x^{n-1}}}-{\frac {(2n-3)b}{2a(n-1)}}\int {\frac {1}{x^{n-1}}}{\sqrt {a+bx}}dx,n\neq 1}