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Senin, 26 November 2012

Integral dari Fungsi rasional



\int (ax + b)^n dx = \frac{(ax + b)^{n+1}}{a(n + 1)} \qquad\mbox{(untuk } n\neq -1\mbox{)}\,\!
\int\frac{1}{ax + b} dx = \frac{1}{a}\ln\left|ax + b\right|
\int x(ax + b)^n dx = \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{)}

\int\frac{x}{ax + b} dx = \frac{x}{a} - \frac{b}{a^2}\ln\left|ax + b\right|
\int\frac{x}{(ax + b)^2} dx = \frac{b}{a^2(ax + b)} + \frac{1}{a^2}\ln\left|ax + b\right|
\int\frac{x}{(ax + b)^n} dx = \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{)}

\int\frac{x^2}{ax + b} dx = \frac{1}{a^3}\left(\frac{(ax + b)^2}{2} - 2b(ax + b) + b^2\ln\left|ax + b\right|\right)
\int\frac{x^2}{(ax + b)^2} dx = \frac{1}{a^3}\left(ax + b - 2b\ln\left|ax + b\right| - \frac{b^2}{ax + b}\right)
\int\frac{x^2}{(ax + b)^3} dx = \frac{1}{a^3}\left(\ln\left|ax + b\right| + \frac{2b}{ax + b} - \frac{b^2}{2(ax + b)^2}\right)
\int\frac{x^2}{(ax + b)^n} dx = \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{)}

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

\int\frac{1}{ax^2+bx+c} dx =
  • \frac{2}{\sqrt{4ac-b^2}}\arctan\frac{2ax+b}{\sqrt{4ac-b^2}} \qquad\mbox{(untuk }4ac-b^2>0\mbox{)}
  • -\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{)}
  • -\frac{2}{2ax+b}\qquad\mbox{(untuk }4ac-b^2=0\mbox{)}
\int\frac{x}{ax^2+bx+c} dx = \frac{1}{2a}\ln\left|ax^2+bx+c\right|-\frac{b}{2a}\int\frac{dx}{ax^2+bx+c}

\int\frac{mx+n}{ax^2+bx+c} dx =
  • \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{)}
  • \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{)}
  • \frac{m}{2a}\ln\left|ax^2+bx+c\right|-\frac{2an-bm}{a(2ax+b)} \,\,\,\,\,\,\,\,\,\, \qquad\mbox{(untuk }4ac-b^2=0\mbox{)}

\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\,\!
\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\,\!
\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 diatas dengan memanfaatkan integrasi parsial, dengan menguraikan fungsi rasional menjadi penjumlahan fungsi-fungsi dalam bentuk
\frac{ex + f}{\left(ax^2+bx+c\right)^n}.


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