integral from x to x^2 of e^{t^4}

解答

\int_{x}^{x^{2}} e^{t^{4}} d t

- \frac{x ^{2} Γ ( \frac{1}{4} , - ( x ^{2} ) ^{4} )}{4 4 - ( x ^{2} ) ^{4}} - (- \frac{x Γ ( \frac{1}{4} , - x ^{4} )}{4 4 - x ^{4}})

求解步骤

\int_{x}^{x^{2}} e^{t^{4}} d t

这是非初等函数积分:

\int e^{x^{a}} d x = - \frac{x Γ ( \frac{1}{a} , - x ^{a} )}{a a - x ^{a}}

= [- \frac{t Γ ( \frac{1}{4} , - t ^{4} )}{4 4 - t ^{4}}]_{x}^{x^{2}}

带入上下限计算后得:

- \frac{x ^{2} Γ ( \frac{1}{4} , - ( x ^{2} ) ^{4} )}{4 4 - ( x ^{2} ) ^{4}} - (- \frac{x Γ ( \frac{1}{4} , - x ^{4} )}{4 4 - x ^{4}})

= - \frac{x ^{2} Γ ( \frac{1}{4} , - ( x ^{2} ) ^{4} )}{4 4 - ( x ^{2} ) ^{4}} - (- \frac{x Γ ( \frac{1}{4} , - x ^{4} )}{4 4 - x ^{4}})

\int_{- 1}^{3} (2 x + 3)^{2} d x

d er i v a t i v e \frac{x - y}{x}

\frac{d}{d x} (\frac{x ^{2} ( x + 3 )}{16 ( x ^{2} - 4 )})

\frac{d y}{d x} = y, y (11) = 3

\int \frac{1}{x ^{2} - x - 6} d x