imported>VismitaRavindra १ जुलाई २०१६ को १५:४१ बजे

2016-07-01T15:41:05Z

नया पृष्ठ

[[File:Reciprocal integral.svg|thumb|एक पारस्परिक समारोह का ग्रफ्]]
[[File:Log-pole-x 1.svg|thumb|Log-pole-x 1]]

नीचे प्रमुख परिमेय फलनों (rational functions) के '''समाकल''' (integrals) दिये गये हैं।

:{|
|<math>\int (ax + b)^n dx</math> || <math> = \frac{(ax + b)^{n+1}}{a(n + 1)} \qquad\mbox{(for } n\neq -1\mbox{)}\,\!</math>
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|<math>\int\frac{1}{ax + b} dx</math> || <math> = \frac{1}{a}\ln\left|ax + b\right|</math>
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|<math>\int x(ax + b)^n dx</math> || <math> = \frac{a(n + 1)x - b}{a^2(n + 1)(n + 2)} (ax + b)^{n+1} \qquad\mbox{(for }n \not\in \{-1, -2\}\mbox{)}</math>
|}

:{|
|<math>\int\frac{x}{ax + b} dx</math> || <math> = \frac{x}{a} - \frac{b}{a^2}\ln\left|ax + b\right|</math>
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|<math>\int\frac{x}{(ax + b)^2} dx</math>||<math> = \frac{b}{a^2(ax + b)} + \frac{1}{a^2}\ln\left|ax + b\right|</math>
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|<math>\int\frac{x}{(ax + b)^n} dx</math>||<math> = \frac{a(1 - n)x - b}{a^2(n - 1)(n - 2)(ax + b)^{n-1}} \qquad\mbox{(for } n\not\in \{1, 2\}\mbox{)}</math>
|}

:{|
|<math>\int\frac{x^2}{ax + b} dx</math>||<math> = \frac{1}{a^3}\left(\frac{(ax + b)^2}{2} - 2b(ax + b) + b^2\ln\left|ax + b\right|\right)</math>
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|<math>\int\frac{x^2}{(ax + b)^2} dx</math>||<math> = \frac{1}{a^3}\left(ax + b - 2b\ln\left|ax + b\right| - \frac{b^2}{ax + b}\right)</math>
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|<math>\int\frac{x^2}{(ax + b)^3} dx</math>||<math> = \frac{1}{a^3}\left(\ln\left|ax + b\right| + \frac{2b}{ax + b} - \frac{b^2}{2(ax + b)^2}\right)</math>
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|<math>\int\frac{x^2}{(ax + b)^n} dx</math>||<math> = \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{(for } n\not\in \{1, 2, 3\}\mbox{)}</math>
|}

:{|
|<math>\int\frac{1}{x(ax + b)} dx</math>||<math> = -\frac{1}{b}\ln\left|\frac{ax+b}{x}\right|</math>
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|<math>\int\frac{1}{x^2(ax+b)} dx</math>||<math> = -\frac{1}{bx} + \frac{a}{b^2}\ln\left|\frac{ax+b}{x}\right|</math>
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|<math>\int\frac{1}{x^2(ax+b)^2} dx</math>||<math> = -a\left(\frac{1}{b^2(ax+b)} + \frac{1}{ab^2x} - \frac{2}{b^3}\ln\left|\frac{ax+b}{x}\right|\right)</math>
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|<math>\int\frac{1}{x^2+a^2} dx</math>||<math> = \frac{1}{a}\arctan\frac{x}{a}\,\!</math>
|-
|<math>\int\frac{1}{x^2-a^2} dx = </math>||
* <math> -\frac{1}{a}\,\mathrm{arctanh}\frac{x}{a} = \frac{1}{2a}\ln\frac{a-x}{a+x} \qquad\mbox{(for }|x| < |a|\mbox{)}\,\!</math>
|-
| ||
* <math> -\frac{1}{a}\,\mathrm{arccoth}\frac{x}{a} = \frac{1}{2a}\ln\frac{x-a}{x+a} \qquad\mbox{(for }|x| > |a|\mbox{)}\,\!</math>
|}

for <math>a\neq 0:</math>
:{|
|<math>\int\frac{1}{ax^2+bx+c} dx =</math>||
* <math> \frac{2}{\sqrt{4ac-b^2}}\arctan\frac{2ax+b}{\sqrt{4ac-b^2}} \qquad\mbox{(for }4ac-b^2>0\mbox{)}</math>
|-
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* <math> -\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{(for }4ac-b^2<0\mbox{)}</math>
|-
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* <math> -\frac{2}{2ax+b}\qquad\mbox{(for }4ac-b^2=0\mbox{)}</math>
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|<math>\int\frac{x}{ax^2+bx+c} dx</math>||<math> = \frac{1}{2a}\ln\left|ax^2+bx+c\right|-\frac{b}{2a}\int\frac{dx}{ax^2+bx+c}</math>
|}

:{|
|<math>\int\frac{mx+n}{ax^2+bx+c} dx = </math> ||
* <math>\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{(for }4ac-b^2>0\mbox{)}</math>
|-
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* <math>\frac{m}{2a}\ln\left|ax^2+bx+c\right|+\frac{2an-bm}{a\sqrt{b^2-4ac}}\,\mathrm{arctanh}\frac{2ax+b}{\sqrt{b^2-4ac}} \qquad\mbox{(for }4ac-b^2<0\mbox{)}</math>
|-
| ||
* <math> \frac{m}{2a}\ln\left|ax^2+bx+c\right|-\frac{2an-bm}{a(2ax+b)} \,\,\,\,\,\,\,\,\,\, \qquad\mbox{(for }4ac-b^2=0\mbox{)}</math>
|}

: <math>\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\,\!</math>
: <math>\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\,\!</math>
: <math>\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</math>

उपरोक्त समीकरणों के साथ '''[[आंशिक भिन्न]] में बदलकर समाकलन''' की विधि का प्रयोग करके किसी भी परिमेय फलन का समाकल निकाला जा सकता है
: <math>\frac{ex + f}{\left(ax^2+bx+c\right)^n}</math>

== इन्हें भी देखें ==

* [[समाकल सूची]] (List of Integrals)

[[श्रेणी:कलन]]
[[श्रेणी:समाकलन]]
[[श्रेणी:गणित]]

परिमेय फलनों के समाकल की सूची - अवतरण इतिहास

imported>VismitaRavindra १ जुलाई २०१६ को १५:४१ बजे