Формулы. Математика
Интегралы от рациональных функций
Интегралы от рациональных функций
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\[
\int x^{n} dx
\]
\[
\int \frac{dx}{x}
\]
\[
\int (ax + b)^{n} dx
\]
\[
\int {dx \over ax + b}
\]
\[
\int {dx \over (ax + b)^2}
\]
\[
\int {xdx \over ax + b}
\]
\[
\int {x^2 dx \over ax + b}
\]
\[
\int {x^n dx \over ax + b}
\]
\[
\int {xdx \over (ax + b)^2}
\]
\[
\int {x^2dx \over (ax + b)^2}
\]
\[
\int {x^ndx \over (ax + b)^2}
\]
\[
\int {x^3dx \over (ax + b)^2}
\]
\[
\int {x^ndx \over (ax + b)^m}
\]
\[
\int {dx \over x(ax + b)}
\]
\[
\int {dx \over x(ax + b)^2}
\]
\[
\int {dx \over x(ax+b)^3}
\]
\[
\int {dx \over x(ax+b)^n}
\]
\[
\int {dx \over x^2(ax+b)^2}
\]
\[
\int {dx \over x^2(ax+b)}
\]
\[
\int {dx \over x^2(ax+b)^3}
\]
\[
\int {dx \over x^3(ax+b)}
\]
\[
\int {dx \over x^n(ax+b)^m}
\]
\[
\int {ax+b \over \alpha x+\beta}dx
\]
\[
\int {(ax+b)^2 \over \alpha x+\beta}dx
\]
\[
\int {ax+b \over (\alpha x+\beta)^2}dx
\]
\[
\int {(ax+b)^2 \over (\alpha x+\beta)^2}dx
\]
\[
\int {(ax+b)^n \over (\alpha x+\beta)^m}dx
\]
\[
\int {dx \over (ax+b)(\alpha x+\beta)}
\]
\[
\int {dx \over (ax+b)(\alpha x+\beta)^2}
\]
\[
\int {xdx \over (ax+b)(\alpha x+\beta)}
\]
\[
\int {xdx \over (x+a)(x+b)}
\]
\[
\int {xdx \over (x+a)(x+b)^2}
\]
\[
\int {dx \over (x+a)^2(x+b)^2}
\]
\[
\int {xdx \over (x+a)^2(x+b)^2}
\]
\[
\int {x^2dx \over (x+a)^2(x+b)^2}
\]
\[
\int {dx \over x(x+a)(x+b)}
\]
\[
\int {dx \over x^2+a^2}
\]
\[
\int {dx \over b^2x^2+a^2}
\]
\[
\int {xdx \over x^2+a^2}
\]
\[
\int {xdx \over (x^2+a^2)^2}
\]
\[
\int {dx \over (x^2+a^2)^2}
\]
\[
\int {dx \over (x^2+a^2)^3}
\]
\[
\int {dx \over (x^2+a^2)^n}
\]
\[
\int {xdx \over (x^2+a^2)^n}
\]
\[
\int {x^2dx \over x^2+a^2}
\]
\[
\int {x^2dx \over (x^2+a^2)^2}
\]
\[
\int {x^3dx \over (x^2+a^2)^2}
\]
\[
\int {x^2dx \over (x^2+a^2)^n}
\]
\[
\int {dx \over x(x^2+a^2)}
\]
\[
\int {dx \over x^2(x^2+a^2)}
\]
\[
\int {dx \over x(x^2+a^2)^2}
\]
\[
\int {dx \over a^2-x^2}
\]
\[
\int {dx \over a^2-b^2x^2}
\]
\[
\int {dx \over (a^2-x^2)^2}
\]
\[
\int {dx \over (a^2-x^2)^n}
\]
\[
\int {xdx \over a^2-x^2}
\]
\[
\int {x^2dx \over a^2-x^2}
\]
\[
\int {xdx \over (a^2-x^2)^2}
\]
\[
\int {x^2dx \over (a^2-x^2)^2}
\]
\[
\int {x^3dx \over a^2-x^2}
\]
\[
\int {x^3dx \over (a^2-x^2)^2}
\]
\[
\int {xdx \over (a^2-x^2)^n}
\]
\[
\int {dx \over x(a^2-x^2)}
\]
\[
\int {dx \over x(a^2-x^2)^2}
\]
\[
\int {dx \over x^2(a^2-x^2)}
\]
\[
\int {dx \over x^2(a^2-x^2)^2}
\]
\[
\int {dx \over x^3+a^3}
\]
\[
\int {dx \over (x^3+a^3)^2}
\]
\[
\int {xdx \over x^3+a^3}
\]
\[
\int {x^2dx \over x^3+a^3}
\]
\[
\int {dx \over x(x^3+a^3)}
\]
\[
\int {dx \over x(x^3+a^3)^2}
\]
\[
\int {dx \over x^4+a^4}
\]
\[
\int {dx \over x^4-a^4}
\]
\[
\int {xdx \over x^4+a^4}
\]
\[
\int {xdx \over x^4-a^4}
\]
\[
\int {dx \over x(x^4 \pm a^4)}
\]
\[
\int {dx \over x^2(x^4 \pm a^4)}
\]
\[
\int {dx \over x^2 \pm x + 1}
\]
\[
\int {dx \over (x^2+px+q)^2}
\]
\[
\int {dx \over x^2+px+q}
\]
\[
\int {xdx \over x^2+px+q}
\]
\[
\int {dx \over x(x^2+px+q)}
\]
\[
\int x^{n} dx = {x^{n+1} \over n + 1}, \quad n \in Z, \quad n \not = -1
\]
\[
\int \frac{dx}{x} = \ln|x|, \quad x \not = 0
\]
\[
\int (ax + b)^{n} dx = {(ax + b)^{n+1} \over a(n + 1)}, \quad n \in Z, \quad n \not = -1, \quad a \not = 0
\]
\[
\int {dx \over ax + b} = \frac{1}{a} \ln|ax + b|, \quad a \not = 0
\]
\[
\int {dx \over (ax + b)^2} = -\frac{1}{a} \frac{1}{ax + b}, \quad a \not = 0
\]
\[
\int {xdx \over ax + b} = \frac{x}{a} - \frac{b}{a^2}\ln|ax + b|, \quad a \not = 0
\]
\[
\int {x^2 dx \over ax + b} = \frac{x^2}{2a} - \frac{bx}{a^2} + \frac{b^2}{a^3}\ln|ax + b|, \quad a \not = 0
\]
\[
\int {x^n dx \over ax + b} = \sum\limits_{k=0}^{n-1} {(-1)^k \, b^k \, x^{n-k} \over (n-k) \, a^{k+1} } + {(-1)^n \, b^n \over a^{n+1}} \ln|ax + b|, \quad a \not = 0
\]
\[
\int {xdx \over (ax + b)^2} = \frac{b}{a^2(ax+b)} + \frac{1}{a^2}\ln|ax + b|, \quad a \not = 0
\]
\[
\int {x^2dx \over (ax + b)^2} = \frac{x}{a^2} - \frac{b^2}{a^3(ax+b)} - \frac{2b}{a^3}\ln|ax + b|, \quad a \not = 0
\]
\[
\int {x^ndx \over (ax + b)^2} = -\frac{x^n}{a(ax+b)} + \frac{n}{a} \int {x^{n-1} dx \over ax+b}, \quad a \not = 0
\]
\[
\int {x^3dx \over (ax + b)^2} = \frac{x^3}{3a} - \frac{bx^2}{2a^2} + \frac{b^2x}{a^3} - \frac{b^3}{a^4}\ln|ax + b|, \quad a \not = 0
\]
\[
\int {x^ndx \over (ax + b)^m} = -\frac{x^n}{(m-1)a(ax+b)^{m-1}} + \frac{n}{(m-1)a} \int {x^{n-1} dx \over (ax+b)^{m-1}}, \quad a, b \not = 0, \quad m \gt 1
\]
\[
\int {dx \over x(ax + b)} = \frac{1}{b} \ln\left|{x \over ax + b}\right|, \quad b \not = 0
\]
\[
\int {dx \over x(ax + b)^2} = \frac{1}{b(ax + b)} + \frac{1}{b^2} \ln\left|{x \over ax + b}\right|, \quad b \not = 0
\]
\[
\int {dx \over x(ax+b)^3} = {1 \over 2b(ax+b)^2} + {1 \over b^2(ax+b)} + \frac{1}{b^3} \ln\left|{x \over ax+b}\right|, \quad b \not = 0
\]
\[
\int {dx \over x(ax+b)^n} = \sum\limits_{k=1}^{n-1} {1 \over k\,b^{n-k}(ax+b)^k} + \frac{1}{b^n}\ln\left|{x \over ax+b}\right|, \quad b \not = 0
\]
\[
\int {dx \over x^2(ax+b)^2} = {a \over b^2(ax+b)} - {1 \over b^2x} - \frac{2a}{b^3} \ln\left|{x \over ax+b}\right|, \quad b \not = 0
\]
\[
\int {dx \over x^2(ax+b)} = - {1 \over bx} - \frac{a}{b^2} \ln\left|{x \over ax+b}\right|, \quad b \not = 0
\]
\[
\int {dx \over x^2(ax+b)^3} = - {a \over 2b^2(ax+b)^2} - {2a \over b^3(ax+b)} - \frac{1}{b^3x} - \frac{3a}{b^4} \ln\left|{x \over ax+b}\right|, \quad b \not = 0
\]
\[
\int {dx \over x^3(ax+b)} = \frac{a}{b^2x} - {1 \over 2bx^2} + \frac{a^2}{b^3} \ln\left|{x \over ax+b}\right|, \quad b \not = 0
\]
\[
\int {dx \over x^n(ax+b)^m} = - {1 \over (n-1)b\,x^{n-1}(ax+b)^{m-1}} - {(n+m-2)a \over b(n-1)} \int {dx \over x^{n-1}(ax+b)^m}, \quad b \not = 0, n \gt 0
\]
\[
\int {ax+b \over \alpha x+\beta}dx = \frac{ax}{\alpha} - {a\beta - \alpha b \over \alpha^2} \ln|\alpha x + \beta|, \quad \alpha \not = 0
\]
\[
\int {(ax+b)^2 \over \alpha x+\beta}dx = \frac{a^2 x^2}{2\alpha} + \frac{ax}{\alpha}\left(b - {a\beta - \alpha b \over \alpha}\right) + {(a\beta - \alpha b)^2 \over \alpha^3} \ln|\alpha x + \beta|, \quad \alpha \not = 0
\]
\[
\int {ax+b \over (\alpha x+\beta)^2}dx = {a\beta - \alpha b \over \alpha^2(\alpha x+\beta)} + \frac{a}{\alpha^2} \ln|\alpha x + \beta|, \quad \alpha \not = 0
\]
\[
\int {(ax+b)^2 \over (\alpha x+\beta)^2}dx = \frac{a^2x}{\alpha^2} - {(a\beta - \alpha b)^2 \over \alpha^3(\alpha x+\beta)} - \frac{2a(a\beta - \alpha b)}{\alpha^3 } \ln|\alpha x + \beta|, \quad \alpha \not = 0
\]
\[
\int {(ax+b)^n \over (\alpha x+\beta)^m}dx = - {(ax+b)^n \over (m-1)\alpha(\alpha x+\beta)^{m-1}} + {na \over (m-1)\alpha} \int {(ax+b)^{n-1} \over (\alpha x+\beta)^{m-1}} dx, \quad \alpha \not = 0, m \gt 1
\]
\[
\int {dx \over (ax+b)(\alpha x+\beta)} = {-1 \over a\beta - \alpha b} \ln\left|{\alpha x+\beta \over ax+b}\right|, \quad a\beta - \alpha b \not = 0
\]
\[
\int {dx \over (ax+b)(\alpha x+\beta)^2} = {1 \over (a\beta - \alpha b)(\alpha x + \beta)} + {a \over (a\beta - \alpha b)^2} \ln\left|{\alpha x+\beta \over ax+b}\right|, \quad a\beta - \alpha b \not = 0
\]
\[
\int {xdx \over (ax+b)(\alpha x+\beta)} = {-b \over a(a\beta - \alpha b)} \ln|ax+b| + {\beta \over \alpha(a\beta - \alpha b)} \ln|\alpha x + \beta|, \quad a\beta - \alpha b \not = 0, a \not = 0,\alpha \not = 0
\]
\[
\int {xdx \over (x+a)(x+b)} = \frac{a}{a-b}\ln|x+a| - \frac{b}{a-b}\ln|x+b|, \quad a \neq b
\]
\[
\int {xdx \over (x+a)(x+b)^2} = {b \over (a-b)(x+b)} - {a \over (a-b)^2} \ln\left|{a+x \over b+x}\right|, \quad a \neq b
\]
\[
\int {dx \over (x+a)^2(x+b)^2} = - \frac{1}{(a-b)^2}\left({1 \over x+a} + {1 \over x+b}\right) + \frac{2}{(a-b)^3}\ln\left|{a+x \over b+x}\right|, \quad a \neq b
\]
\[
\int {xdx \over (x+a)^2(x+b)^2} = \frac{1}{(a-b)^2}\left({a \over x+a} + {b \over x+b}\right) - \frac{a+b}{(a-b)^3}\ln\left|{a+x \over b+x}\right|, \quad a \neq b
\]
\[
\int {x^2dx \over (x+a)^2(x+b)^2} = - \frac{1}{(a-b)^2}\left({a^2 \over x+a} + {b^2 \over x+b}\right) + \frac{2ab}{(a-b)^3}\ln\left|{a+x \over b+x}\right|, \quad a \neq b
\]
\[
\int {dx \over x(x+a)(x+b)} = \frac{1}{ab}\ln|x| + {\ln|x+a| \over a(b-a)} - {\ln|x+b| \over b(b-a)}, \quad a \neq b, a \neq 0, b \neq 0
\]
\[
\int {dx \over x^2+a^2} = \frac1a arctg\frac{x}{a}, \quad a \neq 0
\]
\[
\int {dx \over b^2x^2+a^2} = \frac{1}{ab} arctg\frac{bx}{a}, \quad a \neq 0, b \neq 0
\]
\[
\int {xdx \over x^2+a^2} = \frac12 \ln(x^2+a^2)
\]
\[
\int {xdx \over (x^2+a^2)^2} = - {1 \over 2(x^2+a^2)}
\]
\[
\int {dx \over (x^2+a^2)^2} = {x \over 2a^2(x^2+a^2)} + \frac{1}{2a^3} arctg\frac{x}{a}, \quad a \neq 0
\]
\[
\int {dx \over (x^2+a^2)^3} = {x \over 4a^2(x^2+a^2)^2} + {3x \over 8a^4(x^2+a^2)} + \frac{3}{8a^5} arctg\frac{x}{a}, \quad a \neq 0
\]
\[
\int {dx \over (x^2+a^2)^n} = {x \over 2(n-1)a^2(x^2+a^2)^{n-1}} + {2n-3 \over (2n-2)a^2} \int {dx \over (x^2+a^2)^{n-1}}, \quad a \neq 0, n \gt 1
\]
\[
\int {xdx \over (x^2+a^2)^n} = - {1 \over 2(n-1)(x^2+a^2)^{n-1}}, \quad n \gt 1
\]
\[
\int {x^2dx \over x^2+a^2} = x - a\,\mathtt{arctg}\frac{x}{a}, \quad a \neq 0
\]
\[
\int {x^2dx \over (x^2+a^2)^2} = - {x \over 2(x^2+a^2)} + \frac{1}{2a}\,\mathtt{arctg}\frac{x}{a}, \quad a \neq 0
\]
\[
\int {x^3dx \over (x^2+a^2)^2} = {a^2 \over 2(x^2+a^2)} + \frac12 \ln(x^2+a^2)
\]
\[
\int {x^2dx \over (x^2+a^2)^n} = - {x \over 2(n-1)(x^2+a^2)^{n-1}} + \frac{1}{2(n-1)} \int {dx \over (x^2+a^2)^{n-1}}, \quad a \neq 0, n \gt 1
\]
\[
\int {dx \over x(x^2+a^2)} = \frac{1}{2a^2} \ln{x^2 \over x^2+a^2}, \quad a \neq 0
\]
\[
\int {dx \over x^2(x^2+a^2)} = - {1 \over a^2x} - \frac{1}{a^3}\,\mathtt{arctg}\frac{x}{a}, \quad a \neq 0
\]
\[
\int {dx \over x(x^2+a^2)^2} = {1 \over 2a^2(x^2+a^2)} + \frac{1}{2a^4} \ln{x^2 \over x^2+a^2}, \quad a \neq 0
\]
\[
\int {dx \over a^2-x^2} = \frac{1}{2a} \ln\left|{a+x \over a-x}\right|, \quad a \neq 0
\]
\[
\int {dx \over a^2-b^2x^2} = \frac{1}{2ab} \ln\left|{a+bx \over a-bx}\right|, \quad a \neq 0, b \neq 0
\]
\[
\int {dx \over (a^2-x^2)^2} = {x \over 2a^2(a^2-x^2)} + \frac{1}{4a^3} \ln\left|{a+x \over a-x}\right|, \quad a \neq 0
\]
\[
\int {dx \over (a^2-x^2)^n} = {x \over 2(n-1)a^2(a^2-x^2)^{n-1}} + {2n-3 \over (2n-2)a^2} \int {dx \over (a^2-x^2)^{n-1}}, \quad a \neq 0, n \gt 1
\]
\[
\int {xdx \over a^2-x^2} = - \frac12 \ln|a^2-x^2|
\]
\[
\int {x^2dx \over a^2-x^2} = - x + \frac{a}{2} \ln\left|{a+x \over a-x}\right|
\]
\[
\int {xdx \over (a^2-x^2)^2} = {1 \over 2(a^2-x^2)}
\]
\[
\int {x^2dx \over (a^2-x^2)^2} = {x \over 2(a^2-x^2)} - \frac{1}{4a} \ln\left|{a+x \over a-x}\right|, \quad a \neq 0
\]
\[
\int {x^3dx \over a^2-x^2} = - {x^2 \over 2} - \frac{a^2}{2} \ln|a^2-x^2|
\]
\[
\int {x^3dx \over (a^2-x^2)^2} = {a^2 \over 2(a^2-x^2)} + \frac{1}{2} \ln|a^2-x^2|
\]
\[
\int {xdx \over (a^2-x^2)^n} = {1 \over (2n-2)(a^2-x^2)^{n-1}}, \quad n \gt 1
\]
\[
\int {dx \over x(a^2-x^2)} = \frac{1}{2a^2} \ln\left|{x^2 \over a^2-x^2}\right|, \quad a \neq 0
\]
\[
\int {dx \over x(a^2-x^2)^2} = {1 \over 2a^2(a^2-x^2)} + \frac{1}{2a^4} \ln\left|{x^2 \over a^2-x^2}\right|, \quad a \neq 0
\]
\[
\int {dx \over x^2(a^2-x^2)} = - {1 \over a^2 x} + \frac{1}{2a^3} \ln\left|{a+x \over a-x}\right|, \quad a \neq 0
\]
\[
\int {dx \over x^2(a^2-x^2)^2} = - {1 \over a^4 x} + {x \over 2a^4(a^2-x^2)} + \frac{3}{4a^5} \ln\left|{a+x \over a-x}\right|, \quad a \neq 0
\]
\[
\int {dx \over x^3+a^3} = \frac{1}{6a^2}\ln{(x+a)^2 \over x^2-ax+a^2} + {1 \over a^2 \sqrt 3} \mathtt{arctg}\frac{2x-a}{a \sqrt 3}, \quad a \neq 0
\]
\[
\int {dx \over (x^3+a^3)^2} = {x \over 3a^3(x^3+a^3)} + {2 \over 3a^3} \int {dx \over x^3+a^3}
\]
\[
\int {xdx \over x^3+a^3} = - \frac{1}{6a}\ln{(x+a)^2 \over x^2-ax+a^2} + {1 \over a \sqrt 3} \mathtt{arctg}\frac{2x-a}{a \sqrt 3}, \quad a \neq 0
\]
\[
\int {x^2dx \over x^3+a^3} = \frac13 \ln|x^3+a^3|
\]
\[
\int {dx \over x(x^3+a^3)} = \frac{1}{3a^3} \ln\left|{x^3 \over x^3+a^3}\right|, \quad a \neq 0
\]
\[
\int {dx \over x(x^3+a^3)^2} = {1 \over 3a^3(x^3+a^3)} + \frac{1}{3a^6} \ln\left|{x^3 \over x^3+a^3}\right|, \quad a \neq 0
\]
\[
\int {dx \over x^4+a^4} = \frac{1}{4\sqrt{2}a^3} \ln\left|{x^2+ax\sqrt{2}+a^2 \over x^2-ax\sqrt{2}+a^2}\right|
+ \frac{1}{2\sqrt{2}a^3} \mathtt{arctg}\frac{x\sqrt{2}+a}{a}
+ \frac{1}{2\sqrt{2}a^3} \mathtt{arctg}\frac{x\sqrt{2}-a}{a}, \quad a \neq 0
\]
\[
\int {dx \over x^4-a^4} = \frac{-1}{2a^3} \mathtt{arctg}\frac{x}{a} - \frac{1}{4a^3} \ln\left|{x+a \over x-a}\right|, \quad a \neq 0
\]
\[
\int {xdx \over x^4+a^4} = \frac{1}{2a^2} \mathtt{arctg}\frac{x^2}{a^2}, \quad a \neq 0
\]
\[
\int {xdx \over x^4-a^4} = - \frac{1}{4a^2} \ln\left|{x^2+a^2 \over x^2-a^2}\right|, \quad a \neq 0
\]
\[
\int {dx \over x(x^4 \pm a^4)} = \pm \frac{1}{4a^4} \ln\left|{x^4 \over x^4 \pm a^4}\right|, \quad a \neq 0
\]
\[
\int {dx \over x^2(x^4 \pm a^4)} = \mp \frac{1}{a^4x} \mp \frac{1}{a^4} \int {x^2 dx \over x^4 \pm a^4}, \quad a \neq 0
\]
\[
\int {dx \over x^2 \pm x + 1} = \frac{2}{\sqrt3} \mathtt{arctg}\frac{2x \pm 1}{\sqrt3}
\]
\[
\int {dx \over (x^2+px+q)^2} = {2x+p \over (4q-p^2)(x^2+px+q)} + {2 \over 4q-p^2} \int {dx \over x^2+px+q}
\]
\[
\int {dx \over x^2+px+q} =
\begin{cases}
{2 \over \sqrt{4q-p^2}} \mathtt{arctg}\frac{2x+p}{\sqrt{4q-p^2}}, \quad p^2-4q \lt 0
\\
{1 \over \sqrt{p^2-4q}} \ln\left|{2x+p-\sqrt{p^2-4q} \over 2x+p+\sqrt{p^2-4q}}\right|, \quad p^2-4q \gt 0
\end{cases}
\]
\[
\int {xdx \over x^2+px+q} =
\begin{cases}
\frac12 \ln|x^2+px+q| - {p \over \sqrt{4q-p^2}} \mathtt{arctg}\frac{2x+p}{\sqrt{4q-p^2}}, \quad p^2-4q \lt 0
\\
\frac12 \ln|x^2+px+q| - {p \over 2\sqrt{p^2-4q}} \ln\left|{2x+p-\sqrt{p^2-4q} \over 2x+p+\sqrt{p^2-4q}}\right|, \quad p^2-4q \gt 0
\end{cases}
\]
\[
\int {dx \over x(x^2+px+q)} =
\begin{cases}
\frac{1}{2q} \ln\left|{x^2 \over x^2+px+q}\right| - {p \over q\sqrt{4q-p^2}} \mathtt{arctg}\frac{2x+p}{\sqrt{4q-p^2}}, \quad p^2-4q \lt 0
\\
\frac{1}{2q} \ln\left|{x^2 \over x^2+px+q}\right| - {p \over 2q\sqrt{p^2-4q}} \ln\left|{2x+p-\sqrt{p^2-4q} \over 2x+p+\sqrt{p^2-4q}}\right|, \quad p^2-4q \gt 0
\end{cases}
\]
