Daftar integral dari fungsi irasional: Perbedaan antara revisi
Konten dihapus Konten ditambahkan
Tidak ada ringkasan suntingan |
Tidak ada ringkasan suntingan |
||
Baris 2:
Artikel ini merupakan '''daftar [[integral]] dari [[fungsi irrasional]]. Untuk daftar integral lainnya, lihat [[tabel integral]].
== Integral melibatkan <math>r = \sqrt{x^2+a^2}</math> ==
: <math>\int r \;dx = \frac{1}{2}\left(x r +a^2\,\ln\left(x+r\right)\right)</math><!-- (1.1) [Abramowitz & Stegun p13 3.3.41] + verified by differentiation -->
Baris 56 ⟶ 55:
: <math>\int\frac{dx}{xr} = -\frac{1}{a}\,\sinh^{-1}\frac{a}{x} = -\frac{1}{a}\ln\left|\frac{a+r}{x}\right|</math>
== Integral melibatkan <math>s = \sqrt{x^2-a^2}</math> ==
Anggap <math>(x^2>a^2)</math>, untuk <math>(x^2<a^2)</math>, perhatikan bagian berikutnya:
: <math>\int xs\;dx = \frac{1}{3}s^3</math>
Baris 77 ⟶ 76:
: <math>\int\frac{x\;dx}{s^{2n+1}} = -\frac{1}{(2n-1)s^{2n-1}} </math>
: <math>\int\frac{x^{2m}\;dx}{s^{2n+1}} = -\frac{1}{2n-1}\frac{x^{2m-1}}{s^{2n-1}}+\frac{2m-1}{2n-1}\int\frac{x^{2m-2}\;dx}{s^{2n-1}}
</math>
: <math>\int\frac{x^2\;dx}{s} = \frac{xs}{2}+\frac{a^2}{2}\ln\left|\frac{x+s}{a}\right|</math>
: <math>\int\frac{x^2\;dx}{s^3} = -\frac{x}{s}+\ln\left|\frac{x+s}{a}\right|</math>
: <math>\int\frac{x^4\;dx}{s} = \frac{x^3s}{4}+\frac{3}{8}a^2xs+\frac{3}{8}a^4\ln\left|\frac{x+s}{a}\right| </math>
: <math>\int\frac{x^4\;dx}{s^3} = \frac{xs}{2}-\frac{a^2x}{s}+\frac{3}{2}a^2\ln\left|\frac{x+s}{a}\right| </math>
: <math>\int\frac{x^4\;dx}{s^5} = -\frac{x}{s}-\frac{1}{3}\frac{x^3}{s^3}+\ln\left|\frac{x+s}{a}\right| </math>
= -\frac{x}{s}-\frac{1}{3}\frac{x^3}{s^3}+\ln\left|\frac{x+s}{a}\right| </math>▼
: <math>\int\frac{x^{2m}\;dx}{s^{2n+1}} = (-1)^{n-m}\frac{1}{a^{2(n-m)}}\sum_{i=0}^{n-m-1}\frac{1}{2(m+i)+1}{n-m-1 \choose i}\frac{x^{2(m+i)+1}}{s^{2(m+i)+1}}\qquad\mbox{(}n>m\ge0\mbox{)}</math>
: <math>\int\frac{dx}{s^3} = -\frac{1}{a^2}\frac{x}{s}</math>
: <math>\int\frac{dx}{s^5} = \frac{1}{a^4}\left[\frac{x}{s}-\frac{1}{3}\frac{x^3}{s^3}\right]</math>
: <math>\int\frac{dx}{s^7} = -\frac{1}{a^6}\left[\frac{x}{s}-\frac{2}{3}\frac{x^3}{s^3}+\frac{1}{5}\frac{x^5}{s^5}\right]</math>
▲: <math>\int\frac{dx}{s^9} =
: <math>\int\frac{x^2\;dx}{s^5} = -\frac{1}{a^2}\frac{x^3}{3s^3}</math>
: <math>\int\frac{x^2\;dx}{s^7} = \frac{1}{a^4}\left[\frac{1}{3}\frac{x^3}{s^3}-\frac{1}{5}\frac{x^5}{s^5}\right]</math>
: <math>\int\frac{x^2\;dx}{s^9} = -\frac{1}{a^6}\left[\frac{1}{3}\frac{x^3}{s^3}-\frac{2}{5}\frac{x^5}{s^5}+\frac{1}{7}\frac{x^7}{s^7}\right]</math>
== Integral melibatkan <math>t = \sqrt{a^2-x^2}</math> ==
: <math>\int t \;dx = \frac{1}{2}\left(xt+a^2\arcsin\frac{x}{a}\right) \qquad\mbox{(}|x|\leq|a|\mbox{)}</math><!-- (3.1) [Abramowitz & Stegun p13 3.3.45] -->
Baris 130 ⟶ 118:
: <math>\int t\;dx = \frac{1}{2}\left(xt-\sgn x\,\cosh^{-1}\left|\frac{x}{a}\right|\right) \qquad\mbox{(untuk }|x|\ge|a|\mbox{)}</math>
== Integral melibatkan <math>R = \sqrt{ax^2+bx+c}</math> ==
: <math>\int\frac{dx}{R} = \frac{1}{\sqrt{a}}\ln\left|2\sqrt{a}R+2ax+b\right| \qquad \mbox{(untuk }a>0\mbox{)}</math><!-- (4.1) [Abramowitz & Stegun p13 3.3.33] + verified by differentiation -->
Baris 152 ⟶ 139:
: <math>\int\frac{x}{R^{2n+1}}\;dx = -\frac{1}{(2n-1)aR^{2n-1}}-\frac{b}{2a}\int\frac{dx}{R^{2n+1}}</math><!-- (4.10) [need reference] - verified by differentiation only -->
: <math>\int\frac{dx}{xR} = -\frac{1}{\sqrt{c}}\ln\left(\frac{2\sqrt{c}R+bx+2c}{x}\right)</math><!-- (4.11) [Abramowitz & Stegun p13 implied by 3.3.38 + 3.3.33] + verified by differentiation -->
: <math>\int\frac{dx}{xR} = -\frac{1}{\sqrt{c}}\sinh^{-1}\left(\frac{bx+2c}{|x|\sqrt{4ac-b^2}}\right)</math><!-- (4.11) [Abramowitz & Stegun p13 implied by 3.3.38 + 3.3.34] + verified by differentiation -->
== Integral melibatkan <math>S = \sqrt{ax+b}</math> ==
: <math>\int \frac{dx}{x\sqrt{ax + b}}\, = \,\frac{-2}{\sqrt{b}}\tanh^{-1}{\sqrt{\frac{ax + b}{b}}} </math>
: <math>\int\frac{\sqrt{ax + b}}{x}\,dx\; = \;2\left(\sqrt{ax + b} - \sqrt{b}\tanh^{-1}{\sqrt{\frac{ax + b}{b}}}\right) </math>
: <math>\int\frac{x^n}{\sqrt{ax + b}}\,dx\; = \;\frac{2}{a(2n+1)}
\left(x^{n}\sqrt{ax + b} - bn\int\frac{x^{n-1}}{\sqrt{ax + b}}\,dx \right)</math>
| |||