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受欢迎的微积分 >

integral of (x+9)/(sqrt(1-2x-2x^2))

解答

\int \frac{x + 9}{1 - 2 x - 2 x ^{2}} d x

解答

\frac{1}{2 2} (- - 4 x^{2} - 4 x + 2 + 17 arcsin (\frac{2 x + 1}{3})) + C

求解步骤

\int \frac{x + 9}{1 - 2 x - 2 x ^{2}} d x

对

1 - 2 x - 2 x^{2}

配方:

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

= \int \frac{x + 9}{- 2 ( x + \frac{1}{2} ) ^{2} + \frac{3}{2}} d x

使用换元积分法

= \int \frac{2 u + 17}{2 - 4 u ^{2} + 3} 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 \frac{2 u + 17}{- 4 u ^{2} + 3} d u

乘开

\frac{2 u + 17}{- 4 u ^{2} + 3} : \frac{2 u}{- 4 u ^{2} + 3} + \frac{17}{- 4 u ^{2} + 3}

= \frac{1}{2} \cdot \int \frac{2 u}{- 4 u ^{2} + 3} + \frac{17}{- 4 u ^{2} + 3} 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 \frac{2 u}{- 4 u ^{2} + 3} d u + \int \frac{17}{- 4 u ^{2} + 3} d u)

\int \frac{2 u}{- 4 u ^{2} + 3} d u = - \frac{1}{2} - 4 u^{2} + 3

\int \frac{17}{- 4 u ^{2} + 3} d u = \frac{17}{2} arcsin (\frac{2}{3} u)

= \frac{1}{2} (- \frac{1}{2} - 4 u^{2} + 3 + \frac{17}{2} arcsin (\frac{2}{3} u))

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

代回

= \frac{1}{2} - \frac{1}{2} - 4 (x + \frac{1}{2})^{2} + 3 + \frac{17}{2} arcsin (\frac{2}{3} (x + \frac{1}{2}))

化简

\frac{1}{2} - \frac{1}{2} - 4 (x + \frac{1}{2})^{2} + 3 + \frac{17}{2} arcsin (\frac{2}{3} (x + \frac{1}{2})) : \frac{1}{2 2} (- - 4 x^{2} - 4 x + 2 + 17 arcsin (\frac{2 x + 1}{3}))

= \frac{1}{2 2} (- - 4 x^{2} - 4 x + 2 + 17 arcsin (\frac{2 x + 1}{3}))

解答补常数

= \frac{1}{2 2} (- - 4 x^{2} - 4 x + 2 + 17 arcsin (\frac{2 x + 1}{3})) + C

作图

流行的例子

integral from 0 to 1 of y/(e^{9y)}

\int_{0}^{1} \frac{y}{e ^{9 y}} d y

斜率 y=2+4x^2-2x^3

s l o p e y = 2 + 4 x^{2} - 2 x^{3}

\int - 3 d x

y^{''}+5y=4t^4,y(0)=0,y^'(0)=0

y^{^{''}} + 5 y = 4 t^{4}, y (0) = 0, y^{^{'}} (0) = 0

tangent y=x^4-17x^2+16

t an g e n t y = x^{4} - 17 x^{2} + 16