integral of (x^2+1)/((x+1)^2)e^x

解答

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

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

求解步骤

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

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \int \frac{e ^{x} ( x ^{2} + 1 )}{( x + 1 ) ^{2}} d x

使用分布积分法

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

\int - e^{x} (x + 1) d x = - e^{x} x

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

化简

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

解答补常数

= - \frac{e ^{x} ( x ^{2} + 1 )}{x + 1} + e^{x} x + C

x \to 1 + lim (x^{2} - 4 x + 3)

\frac{\partial}{\partial x} (4 e^{\frac{x ^{2}}{y}} - cos (x y^{3}))

x \to 0 lim (4 x + x)

\int 2 x^{7} + 4 x^{6} + x^{5} + 1 d x

t an g e n t y = 2 x^{3} - 2 x, (- 1, 0)