integral from 0 to 1 of ((x^3+1)/(x+1))

解答

\int_{0}^{1} (\frac{x ^{3} + 1}{x + 1}) d x

\frac{5}{6}

十进制

0.83333 \dots

求解步骤

\int_{0}^{1} \frac{x ^{3} + 1}{x + 1} d x

化简

\frac{x ^{3} + 1}{x + 1} : x^{2} - x + 1

= \int_{0}^{1} x^{2} - x + 1 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_{0}^{1} x^{2} d x - \int_{0}^{1} x d x + \int_{0}^{1} 1 d x

\int_{0}^{1} x^{2} d x = \frac{1}{3}

\int_{0}^{1} x d x = \frac{1}{2}

\int_{0}^{1} 1 d x = 1

= \frac{1}{3} - \frac{1}{2} + 1

化简

\frac{1}{3} - \frac{1}{2} + 1 : \frac{5}{6}

= \frac{5}{6}

\frac{\partial}{\partial x} (x^{\frac{1}{2}} + y^{\frac{1}{2}})

a re a 20 + x - x^{2}, x^{2} - 5 x, [4, 8]

d er i v a t i v e e^{c o s^{2} (π t - 11)}

\frac{d}{d x} (x cos (π x))

\int_{- 5}^{5} \frac{1}{x ^{\frac{2}{3}}} d x