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受欢迎的微积分 >

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

解答

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

解答

8 (- \frac{1}{4 ( 4 x ^{2} + 4 x + 4 )} - \frac{1}{24 3} (2 arctan (\frac{1}{3} (2 x + 1)) + sin (2 arctan (\frac{1}{3} (2 x + 1))))) + C

求解步骤

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

对

x^{2} + x + 1

配方:

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

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

使用换元积分法

= \int \frac{8 ( 2 u - 1 )}{( 4 u ^{2} + 3 ) ^{2}} d u

提出常数:

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

= 8 \cdot \int \frac{2 u - 1}{( 4 u ^{2} + 3 ) ^{2}} d u

乘开

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

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

= 8 (\int \frac{2 u}{( 4 u ^{2} + 3 ) ^{2}} d u - \int \frac{1}{( 4 u ^{2} + 3 ) ^{2}} d u)

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

\int \frac{1}{( 4 u ^{2} + 3 ) ^{2}} d u = \frac{1}{24 3} (2 arctan (\frac{2}{3} u) + sin (2 arctan (\frac{2}{3} u)))

= 8 (- \frac{1}{4 ( 4 u ^{2} + 3 )} - \frac{1}{24 3} (2 arctan (\frac{2}{3} u) + sin (2 arctan (\frac{2}{3} u))))

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

代回

= 8 - \frac{1}{4 ( 4 ( x + \frac{1}{2} ) ^{2} + 3 )} - \frac{1}{24 3} (2 arctan (\frac{2}{3} (x + \frac{1}{2})) + sin (2 arctan (\frac{2}{3} (x + \frac{1}{2}))))

化简

8 - \frac{1}{4 ( 4 ( x + \frac{1}{2} ) ^{2} + 3 )} - \frac{1}{24 3} (2 arctan (\frac{2}{3} (x + \frac{1}{2})) + sin (2 arctan (\frac{2}{3} (x + \frac{1}{2})))) : 8 (- \frac{1}{4 ( 4 x ^{2} + 4 x + 4 )} - \frac{1}{24 3} (2 arctan (\frac{1}{3} (2 x + 1)) + sin (2 arctan (\frac{1}{3} (2 x + 1)))))

= 8 (- \frac{1}{4 ( 4 x ^{2} + 4 x + 4 )} - \frac{1}{24 3} (2 arctan (\frac{1}{3} (2 x + 1)) + sin (2 arctan (\frac{1}{3} (2 x + 1)))))

解答补常数

= 8 (- \frac{1}{4 ( 4 x ^{2} + 4 x + 4 )} - \frac{1}{24 3} (2 arctan (\frac{1}{3} (2 x + 1)) + sin (2 arctan (\frac{1}{3} (2 x + 1))))) + C

作图

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