integral of x^3-6x^2+x+3

解答

\int x^{3} - 6 x^{2} + x + 3 d x

\frac{x ^{4}}{4} - 2 x^{3} + \frac{x ^{2}}{2} + 3 x + C

求解步骤

\int x^{3} - 6 x^{2} + x + 3 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^{3} d x - \int 6 x^{2} d x + \int x d x + \int 3 d x

\int x^{3} d x = \frac{x ^{4}}{4}

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

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

\int 3 d x = 3 x

= \frac{x ^{4}}{4} - 2 x^{3} + \frac{x ^{2}}{2} + 3 x

解答补常数

= \frac{x ^{4}}{4} - 2 x^{3} + \frac{x ^{2}}{2} + 3 x + C

x \to - 7 lim (\frac{x ^{2} - 5}{7 - x})

\int \frac{cos ( x )}{sin ( x ) + cos ( x )} d x

\frac{d}{d x} ((1 + 3 x)^{\frac{1}{2 x}})

\frac{( 3 y ^{2} - t ^{2} )}{y ^{5}} (\frac{d y}{d t}) + \frac{t}{2 y ^{4}} = 0

t an g e n t x, at x = 9