麦克劳林 ln^2(1+x)

解答

麦克劳林

ln^{2} (1 + x)

x^{2} - x^{3} + \frac{11}{12} x^{4} - \frac{5}{6} x^{5} + \frac{137}{180} x^{6} + \dots

求解步骤

应用麦克劳林公式

= 0 + \frac{\frac{d}{d x} ( ln ^{2} ( 1 + x ) ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( ln ^{2} ( 1 + x ) ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( ln ^{2} ( 1 + x ) ) ( 0 )}{3 !} x^{3} + \dots

求导数

= 0 + \frac{0}{1 !} x + \frac{2}{2 !} x^{2} + \frac{- 6}{3 !} x^{3} + \frac{22}{4 !} x^{4} + \frac{- 100}{5 !} x^{5} + \frac{548}{6 !} x^{6} + \dots

整理后得

= x^{2} - x^{3} + \frac{11}{12} x^{4} - \frac{5}{6} x^{5} + \frac{137}{180} x^{6} + \dots

\int \frac{x ^{15}}{1 + x ^{16}} d x

\frac{d}{d x} ((\frac{x - 1}{x + 1})^{2})

\frac{d}{d x} ((6 x - 1) (3 x - 5)^{- 1})

t an g e n t f (x) = cos (\frac{π x}{2}), at x = 0.7

x y d x - (x^{2} - x) d y = 0