integral of e^{-(2r)/a}r^2

解答

\int e^{- \frac{2 r}{a}} r^{2} d r

- \frac{a ( 2 e ^{- \frac{2 r}{a}} r ^{2} - a ^{2} ( - \frac{2 e ^{- \frac{2 r}{a}} r}{a} - e ^{- \frac{2 r}{a}} ) )}{4} + C

求解步骤

\int e^{- \frac{2 r}{a}} r^{2} d r

使用换元积分法

= \int - \frac{a ^{3} e ^{u} u ^{2}}{8} d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{a ^{3}}{8} \cdot \int e^{u} u^{2} d u

使用分布积分法

= - \frac{a ^{3}}{8} (u^{2} e^{u} - \int 2 u e^{u} d u)

\int 2 u e^{u} d u = 2 (e^{u} u - e^{u})

= - \frac{a ^{3}}{8} (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

u = - \frac{2 r}{a}

代回

= - \frac{a ^{3}}{8} ((- \frac{2 r}{a})^{2} e^{- \frac{2 r}{a}} - 2 (e^{- \frac{2 r}{a}} (- \frac{2 r}{a}) - e^{- \frac{2 r}{a}}))

化简

- \frac{a ^{3}}{8} ((- \frac{2 r}{a})^{2} e^{- \frac{2 r}{a}} - 2 (e^{- \frac{2 r}{a}} (- \frac{2 r}{a}) - e^{- \frac{2 r}{a}})) : - \frac{a ( 2 e ^{- \frac{2 r}{a}} r ^{2} - a ^{2} ( - \frac{2 e ^{- \frac{2 r}{a}} r}{a} - e ^{- \frac{2 r}{a}} ) )}{4}

= - \frac{a ( 2 e ^{- \frac{2 r}{a}} r ^{2} - a ^{2} ( - \frac{2 e ^{- \frac{2 r}{a}} r}{a} - e ^{- \frac{2 r}{a}} ) )}{4}

解答补常数

= - \frac{a ( 2 e ^{- \frac{2 r}{a}} r ^{2} - a ^{2} ( - \frac{2 e ^{- \frac{2 r}{a}} r}{a} - e ^{- \frac{2 r}{a}} ) )}{4} + C

t an g e n t f (x) = x^{3}, at - 5

\int - x sin (n x) d x

\frac{d}{d x} (\frac{1}{x} ln (x))

y^{^{''}} - 6 y^{^{'}} + 9 y = 3 x e^{3 x}

\int (cos (3 x)) d x