j 03:21, 22 May 2007 (UTC)
∫ dy y 2 + ρ 2 = 1 2 ln [ y 2 + ρ 2 + y y 2 + ρ 2 − y ] {\displaystyle \int {\frac {\text{dy}}{\sqrt {y^{2}+\rho ^{2}}}}={\frac {1}{2}}\ln \left[{\frac {{\sqrt {y^{2}+\rho ^{2}}}+y}{{\sqrt {y^{2}+\rho ^{2}}}-y}}\right]}
∫ dy y 2 + ρ 2 = ln [ y + y 2 + ρ 2 ] {\displaystyle \int {\frac {\text{dy}}{\sqrt {y^{2}+\rho ^{2}}}}=\ln \left[y+{\sqrt {y^{2}+\rho ^{2}}}\right]}
E = Q 4 π ϵ 0 z R tan − 1 ( R z ) z ^ {\displaystyle E={\frac {Q}{4\pi \epsilon _{0}zR}}\tan ^{-1}\left({\frac {R}{z}}\right){\hat {z}}}
E = ∫ d q z ^ ρ 2 + z 2 , d q = A ρ ρ d ρ d ϕ {\displaystyle E=\int {\frac {dq{\hat {z}}}{\rho ^{2}+z^{2}}}{\text{, }}dq={\frac {A}{\rho }}\rho d\rho d\phi }
( x 2 2 x 3 3 sin ( x ) − cos ( x ) x log ( x ) − x sinh − 1 ( x ) ) {\displaystyle \left({\begin{array}{c}{\frac {x^{2}}{2}}\\{\frac {x^{3}}{3}}\\\sin(x)\\-\cos(x)\\x\log(x)-x\\\sinh ^{-1}(x)\end{array}}\right)}
∫ 0 1 1 ( 1 − x 4 ) 1 / 3 d x // N = 1.16279 {\displaystyle \int _{0}^{1}{\frac {1}{\left(1-x^{4}\right)^{1/3}}}\,dx{\text{//}}N=1.16279} \text{Series}\left[\frac{1}{\left(1-x^4\right)^{1/3}},\{x,0,30\}\right]
1 + x 4 3 + 2 x 8 9 + 14 x 12 81 + 35 x 16 243 + 91 x 20 729 + 728 x 24 6561 + 1976 x 28 19683 + O [ x ] 31 {\displaystyle 1+{\frac {x^{4}}{3}}+{\frac {2x^{8}}{9}}+{\frac {14x^{12}}{81}}+{\frac {35x^{16}}{243}}+{\frac {91x^{20}}{729}}+{\frac {728x^{24}}{6561}}+{\frac {1976x^{28}}{19683}}+O[x]^{31}}