integral of (8t^4+5t^3-6t^2+7t-9)

解答

\int (8 t^{4} + 5 t^{3} - 6 t^{2} + 7 t - 9) d t

\frac{8 t ^{5}}{5} + \frac{5 t ^{4}}{4} - 2 t^{3} + \frac{7 t ^{2}}{2} - 9 t + C

求解步骤

\int 8 t^{4} + 5 t^{3} - 6 t^{2} + 7 t - 9 d t

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 8 t^{4} d t + \int 5 t^{3} d t - \int 6 t^{2} d t + \int 7 t d t - \int 9 d t

\int 8 t^{4} d t = \frac{8 t ^{5}}{5}

\int 5 t^{3} d t = \frac{5 t ^{4}}{4}

\int 6 t^{2} d t = 2 t^{3}

\int 7 t d t = \frac{7 t ^{2}}{2}

\int 9 d t = 9 t

= \frac{8 t ^{5}}{5} + \frac{5 t ^{4}}{4} - 2 t^{3} + \frac{7 t ^{2}}{2} - 9 t

解答补常数

= \frac{8 t ^{5}}{5} + \frac{5 t ^{4}}{4} - 2 t^{3} + \frac{7 t ^{2}}{2} - 9 t + C

\int \frac{3}{x ^{5}} d x

\frac{d}{d x} (x^{3} + a x^{2} y + b x y^{2} + y^{3})

\int e^{0.5 t} d t

x \to 2 lim (\frac{( 3 x - 6 )}{1 - 4 x - 7})

\frac{\partial}{\partial y} (x^{2} y^{3})