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受欢迎的微积分 >

integral of 1/(sqrt((x^2-2x+5)^3))

解答

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

解答

\frac{x - 1}{4 x ^{2} - 2 x + 5} + C

求解步骤

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

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

假定

(x^{2} - 2 x + 5) \geq 0

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

对

x^{2} - 2 x + 5

配方:

(x - 1)^{2} + 4

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

使用换元积分法

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

使用三角换元法

= \int \frac{1}{4 sec ( v )} d v

提出常数:

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

= \frac{1}{4} \cdot \int \frac{1}{sec ( v )} d v

使用三角恒等式改写

= \frac{1}{4} \cdot \int cos (v) d v

使用积分常见定则:

\int cos (v) d v = sin (v)

= \frac{1}{4} sin (v)

代回

= \frac{1}{4} sin (arctan (\frac{1}{2} (x - 1)))

化简

\frac{1}{4} sin (arctan (\frac{1}{2} (x - 1))) : \frac{x - 1}{4 x ^{2} - 2 x + 5}

= \frac{x - 1}{4 x ^{2} - 2 x + 5}

解答补常数

= \frac{x - 1}{4 x ^{2} - 2 x + 5} + C

作图

流行的例子

integral of 1/(sqrt(x)(\sqrt{x)-1)}

\int \frac{1}{x ( x - 1 )} d x

x y^{^{'}} + 2 y = 2 e x

derivative of sec^2(3x)

\frac{d}{d x} (sec^{2} (3 x))

y^{''}-2*y^'=sin(x)

y^{^{''}} - 2 \cdot y^{^{'}} = sin (x)

integral from 0 to 24 of sqrt(9+3x)

\int_{0}^{24} 9 + 3 x d x