integral of (x^3+1)/(x+1)

解答

\int \frac{x ^{3} + 1}{x + 1} d x

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

求解步骤

\int \frac{x ^{3} + 1}{x + 1} d x

化简

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

= \int 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 x^{2} d x - \int x d x + \int 1 d x

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

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

\int 1 d x = x

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

解答补常数

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

x \to 3 - lim (\frac{3}{x - 3})

\frac{d ^{2}}{d x ^{2}} (\frac{3}{4 t ^{6}})

t an g e n t f (x) = \frac{( x - 5 )}{( x - 6 ) , ( 7 , 2 )}

\frac{d}{d x} (12 x^{3} - 12 x^{2} - 24 x)

\int 20 e^{- 20 x} d x