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受欢迎的微积分 >

integral of 1/((x^2+2x+4)^2)

解答

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

解答

\frac{1}{6 3} (arctan (\frac{1}{3} (x + 1)) + \frac{1}{2} sin (2 arctan (\frac{1}{3} (x + 1)))) + C

求解步骤

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

对

x^{2} + 2 x + 4

配方:

(x + 1)^{2} + 3

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

使用换元积分法

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

使用换元积分法

= \int \frac{3}{( 3 v ^{2} + 3 ) ^{2}} d v

提出常数:

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

= 3 \cdot \int \frac{1}{( 3 v ^{2} + 3 ) ^{2}} d v

分解

\frac{1}{( 3 v ^{2} + 3 ) ^{2}}

= 3 \cdot \int \frac{1}{9 ( v ^{2} + 1 ) ^{2}} d v

提出常数:

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

= 3 \frac{1}{9} \cdot \int \frac{1}{( v ^{2} + 1 ) ^{2}} d v

使用三角换元法

= 3 \frac{1}{9} \cdot \int \frac{1}{sec ^{4} ( w ) cos ^{2} ( w )} d w

使用三角恒等式改写

= 3 \frac{1}{9} \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

= 3 \frac{1}{9} \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

= 3 \frac{1}{9} \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)

= 3 \frac{1}{9} \cdot \frac{1}{2} (w + \frac{1}{2} sin (2 w))

代回

= 3 \frac{1}{9} \cdot \frac{1}{2} (arctan (\frac{x + 1}{3}) + \frac{1}{2} sin (2 arctan (\frac{x + 1}{3})))

化简

3 \frac{1}{9} \cdot \frac{1}{2} (arctan (\frac{x + 1}{3}) + \frac{1}{2} sin (2 arctan (\frac{x + 1}{3}))) : \frac{1}{6 3} (arctan (\frac{1}{3} (x + 1)) + \frac{1}{2} sin (2 arctan (\frac{1}{3} (x + 1))))

= \frac{1}{6 3} (arctan (\frac{1}{3} (x + 1)) + \frac{1}{2} sin (2 arctan (\frac{1}{3} (x + 1))))

解答补常数

= \frac{1}{6 3} (arctan (\frac{1}{3} (x + 1)) + \frac{1}{2} sin (2 arctan (\frac{1}{3} (x + 1)))) + C

作图

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