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受欢迎的微积分 >

integral of sqrt(x^6-x^8)

解答

\int x^{6} - x^{8} d x

解答

- \frac{1}{15} (3 x^{2} + 2) (- x^{2} + 1)^{\frac{3}{2}} + C

求解步骤

\int x^{6} - x^{8} d x

化简

x^{6} - x^{8} : x^{3} 1 - x^{2}

= \int x^{3} 1 - x^{2} d x

使用换元积分法

= \int - \frac{( - u + 1 ) u}{2} d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{2} \cdot \int (- u + 1) u d u

乘开

(- u + 1) u : - u^{\frac{3}{2}} + u

= - \frac{1}{2} \cdot \int - u^{\frac{3}{2}} + u d u

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

= - \frac{1}{2} (- \int u^{\frac{3}{2}} d u + \int u d u)

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

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

= - \frac{1}{2} (- \frac{2}{5} u^{\frac{5}{2}} + \frac{2}{3} u^{\frac{3}{2}})

u = 1 - x^{2}

代回

= - \frac{1}{2} (- \frac{2}{5} (1 - x^{2})^{\frac{5}{2}} + \frac{2}{3} (1 - x^{2})^{\frac{3}{2}})

化简

- \frac{1}{2} (- \frac{2}{5} (1 - x^{2})^{\frac{5}{2}} + \frac{2}{3} (1 - x^{2})^{\frac{3}{2}}) : - \frac{1}{15} (3 x^{2} + 2) (- x^{2} + 1)^{\frac{3}{2}}

= - \frac{1}{15} (3 x^{2} + 2) (- x^{2} + 1)^{\frac{3}{2}}

解答补常数

= - \frac{1}{15} (3 x^{2} + 2) (- x^{2} + 1)^{\frac{3}{2}} + C

作图

流行的例子

tangent f(x)=sin(sin(x)),(2pi,0)

t an g e n t f (x) = sin (sin (x)), (2 π, 0)

integral of (Inx)/x

\int \frac{I n x}{x} d x

integral of (12)/(sec(2x)-1)

\int \frac{12}{sec ( 2 x ) - 1} d x

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

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

y^'-y=e^{-t}sin(t)

y^{^{'}} - y = e^{- t} sin (t)