integral of x^2+4e^x+C

解答

\int x^{2} + 4 e^{x} + C d x

\frac{x ^{3}}{3} + 4 e^{x} + C x + C

求解步骤

\int x^{2} + 4 e^{x} + C d x

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 x^{2} d x + \int 4 e^{x} d x + \int C d x

\int x^{2} d x = \frac{x ^{3}}{3}

\int 4 e^{x} d x = 4 e^{x}

\int C d x = C x

= \frac{x ^{3}}{3} + 4 e^{x} + C x

解答补常数

= \frac{x ^{3}}{3} + 4 e^{x} + C x + C

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

\int_{0}^{1} cos (nπ x) d x

\frac{d}{d x} (ln (6 x^{4} - 9 x - 3))

y^{^{''}} + 8 y = 0

\frac{d}{d x} (arctan (9^{x}))