integral of ((x^2+1)/(x^2))

解答

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

- \frac{1}{x} + x + C

求解步骤

\int \frac{x ^{2} + 1}{x ^{2}} d x

\frac{x ^{2} + 1}{x ^{2}} = \frac{1}{x ^{2}} + 1

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

使用指数法则:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{x ^{2}} = x^{- 2}

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

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

\int 1 d x = x

= - \frac{1}{x} + x

解答补常数

= - \frac{1}{x} + x + C

\frac{d}{d x} (\frac{( x - 4 )}{x})

\frac{\partial}{\partial x} (6 x sin (8 x^{2} y))

d er i v a t i v e - \frac{2 x}{( 1 + x ^{2} ) ^{2}}

\frac{d}{d x} ((x^{2} + 2 x)^{5})

a re a y = x^{2}, y = x, y = 2