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受欢迎的微积分 >

integral of 1/((x^2-6x+11)^2)

解答

\int \frac{1}{( x ^{2} - 6 x + 11 ) ^{2}} d x

解答

\frac{1}{4 2} (arctan (\frac{1}{2} (x - 3)) + \frac{1}{2} sin (2 arctan (\frac{1}{2} (x - 3)))) + C

求解步骤

\int \frac{1}{( x ^{2} - 6 x + 11 ) ^{2}} d x

对

x^{2} - 6 x + 11

配方:

(x - 3)^{2} + 2

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

使用换元积分法

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

使用换元积分法

= \int \frac{2}{( 2 v ^{2} + 2 ) ^{2}} d v

提出常数:

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

= 2 \cdot \int \frac{1}{( 2 v ^{2} + 2 ) ^{2}} d v

分解

\frac{1}{( 2 v ^{2} + 2 ) ^{2}}

= 2 \cdot \int \frac{1}{4 ( v ^{2} + 1 ) ^{2}} d v

提出常数:

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

= 2 \frac{1}{4} \cdot \int \frac{1}{( v ^{2} + 1 ) ^{2}} d v

使用三角换元法

= 2 \frac{1}{4} \cdot \int \frac{1}{sec ^{4} ( w ) cos ^{2} ( w )} d w

使用三角恒等式改写

= 2 \frac{1}{4} \cdot \int \frac{1 + cos ( 2 w )}{2} d w

提出常数:

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

= 2 \frac{1}{4} \cdot \frac{1}{2} \cdot \int 1 + cos (2 w) d w

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

= 2 \frac{1}{4} \cdot \frac{1}{2} (\int 1 d w + \int cos (2 w) d w)

\int 1 d w = w

\int cos (2 w) d w = \frac{1}{2} sin (2 w)

= 2 \frac{1}{4} \cdot \frac{1}{2} (w + \frac{1}{2} sin (2 w))

代回

= 2 \frac{1}{4} \cdot \frac{1}{2} (arctan (\frac{x - 3}{2}) + \frac{1}{2} sin (2 arctan (\frac{x - 3}{2})))

化简

2 \frac{1}{4} \cdot \frac{1}{2} (arctan (\frac{x - 3}{2}) + \frac{1}{2} sin (2 arctan (\frac{x - 3}{2}))) : \frac{1}{4 2} (arctan (\frac{1}{2} (x - 3)) + \frac{1}{2} sin (2 arctan (\frac{1}{2} (x - 3))))

= \frac{1}{4 2} (arctan (\frac{1}{2} (x - 3)) + \frac{1}{2} sin (2 arctan (\frac{1}{2} (x - 3))))

解答补常数

= \frac{1}{4 2} (arctan (\frac{1}{2} (x - 3)) + \frac{1}{2} sin (2 arctan (\frac{1}{2} (x - 3)))) + C

作图

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