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

解答

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

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

求解步骤

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

使用分布积分法

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

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

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

化简

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

解答补常数

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

\int x \cdot arccot (x) d x

\int \frac{( x ^{5} - 1 )}{( x - 1 )} d x

\int \frac{2 x - 3}{x ^{2} - 8 x + 7} d x

\int t t - 6 d t

\int \frac{x}{( x ^{2} + 9 )} d x