麦克劳林 e^{x^2+x}

解答

麦克劳林

e^{x^{2} + x}

1 + x + \frac{3}{2} x^{2} + \frac{7}{6} x^{3} + \frac{25}{24} x^{4} + \dots

求解步骤

应用麦克劳林公式

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

求导数

= 1 + \frac{1}{1 !} x + \frac{3}{2 !} x^{2} + \frac{7}{3 !} x^{3} + \frac{25}{4 !} x^{4} + \dots

整理后得

= 1 + x + \frac{3}{2} x^{2} + \frac{7}{6} x^{3} + \frac{25}{24} x^{4} + \dots

\frac{d}{d x} (3 x^{2} y + y^{3} - 3 x^{2} - 3 y^{2} + 2)

\int \frac{3}{2 x + 7} d x

\int_{1}^{e} (\frac{1}{x}) d x

\frac{d}{d x} (4 e^{- 4 x})

\int 3 (2 t + 5)^{3} d t