integral from 1 to 3 of (e^xln(x))

解答

\int_{1}^{3} (e^{x} ln (x)) d x

e^{3} ln (3) - Ei (3) + Ei (1)

求解步骤

\int_{1}^{3} e^{x} ln (x) d x

使用分布积分法

= [e^{x} ln (x) - \int \frac{e ^{x}}{x} d x]_{1}^{3}

\int \frac{e ^{x}}{x} d x = Ei (x)

= [e^{x} ln (x) - Ei (x)]_{1}^{3}

带入上下限计算后得:

e^{3} ln (3) - Ei (3) + Ei (1)

= e^{3} ln (3) - Ei (3) + Ei (1)

\int_{0}^{\frac{π}{6}} tan (2 θ) d θ

\int_{1}^{2} V d V

\int_{x^{2}}^{1} 4 x d x

\int_{- \infty}^{0} e^{- 5 x} d x

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