ibookSharing Jack: 2010

Integrals involving $r = \sqrt{x^2+a^2}$

$\int r \;dx = \frac{1}{2}\left(x r +a^2\,\ln\left(x+r\right)\right)$

$\int r^3 \;dx = \frac{1}{4}xr^3+\frac{3}{8}a^2xr+\frac{3}{8}a^4\ln\left(x+r\right)$

$\int r^5 \; dx = \frac{1}{6}xr^5+\frac{5}{24}a^2xr^3+\frac{5}{16}a^4xr+\frac{5}{16}a^6\ln\left(x+r\right)$

$\int x r\;dx=\frac{r^3}{3}$

$\int x r^3\;dx=\frac{r^5}{5}$

$\int x r^{2n+1}\;dx=\frac{r^{2n+3}}{2n+3}$

$\int x^2 r\;dx= \frac{xr^3}{4}-\frac{a^2xr}{8}-\frac{a^4}{8}\ln\left(x+r\right)$

$\int x^2 r^3\;dx= \frac{xr^5}{6}-\frac{a^2xr^3}{24}-\frac{a^4xr}{16}-\frac{a^6}{16}\ln\left(x+r\right)$

$\int x^3 r \; dx = \frac{r^5}{5} - \frac{a^2 r^3}{3}$

$\int x^3 r^3 \; dx = \frac{r^7}{7}-\frac{a^2r^5}{5}$

$\int x^3 r^{2n+1} \; dx = \frac{r^{2n+5}}{2n+5} - \frac{a^3 r^{2n+3}}{2n+3}$

$\int x^4 r\;dx= \frac{x^3r^3}{6}-\frac{a^2xr^3}{8}+\frac{a^4xr}{16}+\frac{a^6}{16}\ln\left(x+r\right)$

$\int x^4 r^3\;dx= \frac{x^3r^5}{8}-\frac{a^2xr^5}{16}+\frac{a^4xr^3}{64}+\frac{3a^6xr}{128}+\frac{3a^8}{128}\ln\left(x+r\right)$

$\int x^5 r \; dx = \frac{r^7}{7} - \frac{2 a^2 r^5}{5} + \frac{a^4 r^3}{3}$

$\int x^5 r^3 \; dx = \frac{r^9}{9} - \frac{2 a^2 r^7}{7} + \frac{a^4 r^5}{5}$

$\int x^5 r^{2n+1} \; dx = \frac{r^{2n+7}}{2n+7} - \frac{2a^2r^{2n+5}}{2n+5}+\frac{a^4 r^{2n+3}}{2n+3}$

$\int\frac{r\;dx}{x} = r-a\ln\left|\frac{a+r}{x}\right| = r - a\, \operatorname{arsinh}\frac{a}{x}$

$\int\frac{r^3\;dx}{x} = \frac{r^3}{3}+a^2r-a^3\ln\left|\frac{a+r}{x}\right|$

$\int\frac{r^5\;dx}{x} = \frac{r^5}{5}+\frac{a^2r^3}{3}+a^4r-a^5\ln\left|\frac{a+r}{x}\right|$

$\int\frac{r^7\;dx}{x} = \frac{r^7}{7}+\frac{a^2r^5}{5}+\frac{a^4r^3}{3}+a^6r-a^7\ln\left|\frac{a+r}{x}\right|$

$\int\frac{dx}{r} = \operatorname{arsinh}\frac{x}{a} = \ln\left( \frac{x+r}{a} \right)$

$\int\frac{dx}{r^3} = \frac{x}{a^2r}$

$\int\frac{x\,dx}{r} = r$

$\int\frac{x\,dx}{r^3} = -\frac{1}{r}$

$\int\frac{x^2\;dx}{r} = \frac{x}{2}r-\frac{a^2}{2}\,\operatorname{arsinh}\frac{x}{a} = \frac{x}{2}r-\frac{a^2}{2}\ln\left( \frac {x+r}{a} \right)$

$\int\frac{dx}{xr} = -\frac{1}{a}\,\operatorname{arsinh}\frac{a}{x} = -\frac{1}{a}\ln\left|\frac{a+r}{x}\right|$

[edit] Integrals involving $s = \sqrt{x^2-a^2}$

Assume $(x 2 > a 2)$ , for $(x 2 < a 2)$ , see next section:

$\int s\;dx = \frac{1}{2}\left( xs-a^{2}\ln(x+s)\right)$

$\int xs\;dx = \frac{1}{3}s^3$

$\int\frac{s\;dx}{x} = s - a\arccos\left|\frac{a}{x}\right|$

$\int\frac{dx}{s} = \ln\left|\frac{x+s}{a}\right|$

Here $\ln\left|\frac{x+s}{a}\right| =\mathrm{sgn}(x)\,\operatorname{arcosh}\left|\frac{x}{a}\right| =\frac{1}{2}\ln\left(\frac{x+s}{x-s}\right)$ , where the positive value of $\operatorname{arcosh}\left|\frac{x}{a}\right|$ is to be taken.

$\int\frac{x\;dx}{s} = s$

$\int\frac{x\;dx}{s^3} = -\frac{1}{s}$

$\int\frac{x\;dx}{s^5} = -\frac{1}{3s^3}$

$\int\frac{x\;dx}{s^7} = -\frac{1}{5s^5}$

$\int\frac{x\;dx}{s^{2n+1}} = -\frac{1}{(2n-1)s^{2n-1}}$

$\int\frac{x^{2m}\;dx}{s^{2n+1}} = -\frac{1}{2n-1}\frac{x^{2m-1}}{s^{2n-1}}+\frac{2m-1}{2n-1}\int\frac{x^{2m-2}\;dx}{s^{2n-1}}$

$\int\frac{x^2\;dx}{s} = \frac{xs}{2}+\frac{a^2}{2}\ln\left|\frac{x+s}{a}\right|$

$\int\frac{x^2\;dx}{s^3} = -\frac{x}{s}+\ln\left|\frac{x+s}{a}\right|$

$\int\frac{x^4\;dx}{s} = \frac{x^3s}{4}+\frac{3}{8}a^2xs+\frac{3}{8}a^4\ln\left|\frac{x+s}{a}\right|$

$\int\frac{x^4\;dx}{s^3} = \frac{xs}{2}-\frac{a^2x}{s}+\frac{3}{2}a^2\ln\left|\frac{x+s}{a}\right|$

$\int\frac{x^4\;dx}{s^5} = -\frac{x}{s}-\frac{1}{3}\frac{x^3}{s^3}+\ln\left|\frac{x+s}{a}\right|$

$\text{[math]}$ m\ge0\mbox{)}" src="http://upload.wikimedia.org/math/7/e/c/7ec61e0b86221a7e14575832d055c1d9.png">

$\int\frac{dx}{s^3}=-\frac{1}{a^2}\frac{x}{s}$

$\int\frac{dx}{s^5}=\frac{1}{a^4}\left[\frac{x}{s}-\frac{1}{3}\frac{x^3}{s^3}\right]$

$\int\frac{dx}{s^7} =-\frac{1}{a^6}\left[\frac{x}{s}-\frac{2}{3}\frac{x^3}{s^3}+\frac{1}{5}\frac{x^5}{s^5}\right]$

$\int\frac{dx}{s^9} =\frac{1}{a^8}\left[\frac{x}{s}-\frac{3}{3}\frac{x^3}{s^3}+\frac{3}{5}\frac{x^5}{s^5}-\frac{1}{7}\frac{x^7}{s^7}\right]$

$\int\frac{x^2\;dx}{s^5}=-\frac{1}{a^2}\frac{x^3}{3s^3}$

$\int\frac{x^2\;dx}{s^7} = \frac{1}{a^4}\left[\frac{1}{3}\frac{x^3}{s^3}-\frac{1}{5}\frac{x^5}{s^5}\right]$

$\int\frac{x^2\;dx}{s^9} = -\frac{1}{a^6}\left[\frac{1}{3}\frac{x^3}{s^3}-\frac{2}{5}\frac{x^5}{s^5}+\frac{1}{7}\frac{x^7}{s^7}\right]$

[edit] Integrals involving $u = \sqrt{a^2-x^2}$

$\int u \;dx = \frac{1}{2}\left(xu+a^2\arcsin\frac{x}{a}\right) \qquad\mbox{(}|x|\leq|a|\mbox{)}$

$\int xu\;dx = -\frac{1}{3} u^3 \qquad\mbox{(}|x|\leq|a|\mbox{)}$

$\int x^2u\;dx = -\frac{x}{4} u^3+\frac{a^2}{8}(xu+a^2\arcsin\frac{x}{a}) \qquad\mbox{(}|x|\leq|a|\mbox{)}$

$\int\frac{u\;dx}{x} = u-a\ln\left|\frac{a+u}{x}\right| \qquad\mbox{(}|x|\leq|a|\mbox{)}$

$\int\frac{dx}{u} = \arcsin\frac{x}{a} \qquad\mbox{(}|x|\leq|a|\mbox{)}$

$\int\frac{x^2\;dx}{u} = \frac{1}{2}\left(-xu+a^2\arcsin\frac{x}{a}\right) \qquad\mbox{(}|x|\leq|a|\mbox{)}$

$\int u\;dx = \frac{1}{2}\left(xu-\sgn x\,\operatorname{arcosh}\left|\frac{x}{a}\right|\right) \qquad\mbox{(for }|x|\ge|a|\mbox{)}$

$\int \frac{x}{u}\;dx = -u \qquad\mbox{(}|x|\leq|a|\mbox{)}$

[edit] Integrals involving $R = \sqrt{ax^2+bx+c}$

Assume (ax² + bx + c) cannot be reduced to the following expression (px + q)² for some p and q.

$\text{[math]}$ 0\mbox{)}" src="http://upload.wikimedia.org/math/b/9/7/b9786694b8364705405868b754f3e73a.png">

$\text{[math]}$ 0\mbox{, }4ac-b^2>0\mbox{)}" src="http://upload.wikimedia.org/math/0/d/6/0d6ddd4644254d630d3993720801c6ad.png">

$\text{[math]}$ 0\mbox{, }4ac-b^2=0\mbox{)}" src="http://upload.wikimedia.org/math/6/c/f/6cfab7115ce8ce8a82a1d238a1a80b7b.png">

$\int\frac{dx}{R} = -\frac{1}{\sqrt{-a}}\arcsin\frac{2ax+b}{\sqrt{b^2-4ac}} \qquad \mbox{(for }a<0\mbox{, }4ac-b^2<0\mbox{, }\left|2ax+b\right|<\sqrt{b^2-4ac}\mbox{)}$

$\int\frac{dx}{R^3} = \frac{4ax+2b}{(4ac-b^2)R}$

$\int\frac{dx}{R^5} = \frac{4ax+2b}{3(4ac-b^2)R}\left(\frac{1}{R^2}+\frac{8a}{4ac-b^2}\right)$

$\int\frac{dx}{R^{2n+1}} = \frac{2}{(2n-1)(4ac-b^2)}\left(\frac{2ax+b}{R^{2n-1}}+4a(n-1)\int\frac{dx}{R^{2n-1}}\right)$

$\int\frac{x}{R}\;dx = \frac{R}{a}-\frac{b}{2a}\int\frac{dx}{R}$

$\int\frac{x}{R^3}\;dx = -\frac{2bx+4c}{(4ac-b^2)R}$

$\int\frac{x}{R^{2n+1}}\;dx = -\frac{1}{(2n-1)aR^{2n-1}}-\frac{b}{2a}\int\frac{dx}{R^{2n+1}}$

$\int\frac{dx}{xR}=-\frac{1}{\sqrt{c}}\ln\left(\frac{2\sqrt{c}R+bx+2c}{x}\right)$

$\int\frac{dx}{xR}=-\frac{1}{\sqrt{c}}\operatorname{arsinh}\left(\frac{bx+2c}{|x|\sqrt{4ac-b^2}}\right)$

[edit] Integrals involving $S = \sqrt{ax+b}$

$\int S {dx} = \frac{2 S^{3}}{3 a}$

$\int \frac{dx}{S} = \frac{2S}{a}$

$\text{[math]}$ 0, \quad a x > 0\mbox{)} \\ -\frac{2}{\sqrt{b}} \mathrm{artanh}\left( \frac{S}{\sqrt{b}}\right) & \mbox{(for }b > 0, \quad a x < src="http://upload.wikimedia.org/math/1/e/b/1ebd0028aa19fcf88af1c1ac87e293b9.png">

$\text{[math]}$ 0, \quad a x > 0\mbox{)} \\ 2 \left( S - \sqrt{b}\,\mathrm{artanh}\left( \frac{S}{\sqrt{b}}\right)\right) & \mbox{(for }b > 0, \quad a x < src="http://upload.wikimedia.org/math/6/1/2/612fd00fa1ae63401b55deaa24ee002f.png">

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

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

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

Code	Region
AB	Egypt, Saudi Arabia, United Arab Emirates
B	Ireland, UK, also used for some replacement phones
C	Canada
CZ	Czech Republic
DN	Austria, Germany, Netherlands
E	Mexico
EE	Estonia
FB	France, Luxembourg
FD	Austria, Liechtenstein, Switzerland
GR	Greece
HN	India
IP	Italy
J	Japan
KN	Norway
KS	Finland, Sweden
LA	Colombia, Ecuador, El Salvador, Guatemala, Honduras, Peru
LE	Argentina
LL	USA
LZ	Chile, Paraguay, Uruguay
MG	Hungary
NF	Belgium, France, Luxembourg
PL	Poland
PO	Portugal
PP	Philippines
RO	Romania
RS	Russia
SL	Slovakia
SO	South Africa
T	Italy
TA	Taiwan
TU	Turkey
X	Australia, New Zealand
Y	Spain
ZA	Singapore
ZP	Hong Kong, Macau

ibookSharing Jack

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GO TO CMD by Run as Administrator :

`1.cd C:\WINDOWS\SYSTEM32\ 2.diskpart (if your version is 4.0 , u can't do it , it should be uper than 4.0) 3.list disk 4.Select disk x ( if your flash drive is disk x ) 5.clean 6.create partition primary 7.Select partition 1 8.active 9.format fs=ntfs 10.(wait until it 100 persent completed).`

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Formula of integrals of irrational functions

Integrals involving $r = \sqrt{x^2+a^2}$

[edit] Integrals involving $s = \sqrt{x^2-a^2}$

[edit] Integrals involving $u = \sqrt{a^2-x^2}$

[edit] Integrals involving $R = \sqrt{ax^2+bx+c}$

[edit] Integrals involving $S = \sqrt{ax+b}$