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受欢迎的微积分 >

integral of 1/((-u^2+u-1))

解答

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

解答

- \frac{2}{3} arctan (\frac{2 u - 1}{3}) + C

求解步骤

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

因式分解

- u^{2} + u - 1 : - (u^{2} - u + 1)

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

提出常数:

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

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

对

u^{2} - u + 1

配方:

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

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

使用换元积分法

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

提出常数:

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

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

使用换元积分法

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

提出常数:

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

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

使用积分常见定则:

\int \frac{1}{w ^{2} + 1} d w = arctan (w)

= - 4 \cdot \frac{1}{2 3} arctan (w)

代回

= - 4 \cdot \frac{1}{2 3} arctan (\frac{2}{3} (u - \frac{1}{2}))

化简

- 4 \cdot \frac{1}{2 3} arctan (\frac{2}{3} (u - \frac{1}{2})) : - \frac{2}{3} arctan (\frac{2 u - 1}{3})

= - \frac{2}{3} arctan (\frac{2 u - 1}{3})

解答补常数

= - \frac{2}{3} arctan (\frac{2 u - 1}{3}) + C

作图

流行的例子

derivative of e^{cos(x}+cos(e^x))

\frac{d}{d x} (e^{c o s (x)} + cos (e^{x}))

derivative (1+cos^2(x))^3

d er i v a t i v e (1 + cos^{2} (x))^{3}

(1+x^2)(dy)/(dx)=2xy+1

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

derivative of 2x^2sqrt(2-x)

\frac{d}{d x} (2 x^{2} 2 - x)

derivative of (16/(x^2))

\frac{d}{d x} (\frac{16}{x ^{2}})