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= ∫[0 ~ 1]2x(1 + x²)⁴dx= ∫[0 ~1](1 + u)⁴(2x dx) <-(令 u = x²)= ∫[0 ~1](1 + u)⁴du <-(du = 2x)= ⅕(1 + u)⁵[0 ~ 1]= ⅕(1 + x²)⁵[0 ~ 1]= ⅕[(1 + 1)⁵ - (1 + 0...

= ∫(3x⁴ + 6/√x)dx= ∫(3x⁴ + 6x^-½)dx= 3x⁴⁺¹/(4 +1) + 6x^(-½ + 1)/(-½ + 1)= ⅗x⁵ + 6x^½/(1/2)= ⅗x⁵ + 2√x 希望幫到你，不明白的話可再補充問題！ 2012-07-05 12:33:58 補充...

y³ + xy + x² = 03y²y' +xy' + y + 2x = 0(x + 3y²)y' =- 2x - yy' = - (2x + y)/(x + 3y²) =切線斜率= - [2(- 2) + 2]/[-2 + 3(2)²]= ⅕ 重要公式：f'(uv) = uf'(v)+ vf'(u) 希望幫到你，不明白的話可再補充問題！ 2012-07-03 22:06:31 補充...

3x²z + y³ - xyz³ = 03[x²(∂z/∂x) + z(2x)] - y[x(3z²)(∂z/∂x) + z³] = 03x²(∂z/∂x) + 6xz - 3xyz²(∂z/∂x) - yz³ = 0(3x² -3xyz²)(∂z/∂x) = yz³ - 6xz∂z/∂x = (yz³ - 6xz)/(3x²...

= x⁷ - x⁴ - x² - x – 1= x⁴(x³ - 1) – (x² + x + 1)= x⁴(x – 1)(x² + x + 1) - (x² + x + 1)= (x² + x + 1)(x⁵ - x⁴ - 1)= (x² + x + 1)[x⁵ - x⁴ + x³ - (x³ + 1...

圖片參考：http://i1099.photobucket.com/albums/g395/jasoncube/672A547D540Dh-1_zpsdfe48daa.png http://i1099.photobucket.com/albums/g395/jasoncube/672A547D540Dh-1_zpsdfe48daa.png 2013-02-15 00:32:13 補充： http://hk.knowledge.yahoo.com/question/article?qid=6912021400443

令 u = x², du =2xdx.令 dv = eˣdx, v = eˣ.∴ ∫x²eˣdx=x²eˣ - ∫eˣ(2x dx)=x²eˣ - 2∫x eˣdx 令 p = x, dp = dx令 dq = eˣdx, q = eˣ.∴ ∫x eˣdx=x eˣ - ∫eˣdx=x eˣ - eˣ...

3x²z + y³ - xyz = 03[x²(∂z/∂x) + z(2x)] + 0 - y[x(3z²(∂z/∂x) + z³] = 03x²(∂z/∂x) + 6xz - 3xyz²(∂z/∂x) - yz³ = 03x(x - yz²)(∂z/∂x) = yz³ - 6xz∂z/∂x = z(yz² - 6x)/[3x(x - yz²)] 希望幫到你，不明白...