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受欢迎的微积分 >

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

解答

\int \frac{x}{x ^{2} + x + 2} d x

解答

x^{2} + x + 2 - \frac{1}{2} ln \frac{2}{7} x + \frac{1}{7} + \frac{4 x ^{2} + 4 x + 8}{7} + C

求解步骤

\int \frac{x}{x ^{2} + x + 2} d x

对

x^{2} + x + 2

配方:

(x + \frac{1}{2})^{2} + \frac{7}{4}

= \int \frac{x}{( x + \frac{1}{2} ) ^{2} + \frac{7}{4}} d x

使用换元积分法

= \int \frac{2 u - 1}{4 u ^{2} + 7} d u

乘开

\frac{2 u - 1}{4 u ^{2} + 7} : \frac{2 u}{4 u ^{2} + 7} - \frac{1}{4 u ^{2} + 7}

= \int \frac{2 u}{4 u ^{2} + 7} - \frac{1}{4 u ^{2} + 7} d u

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

= \int \frac{2 u}{4 u ^{2} + 7} d u - \int \frac{1}{4 u ^{2} + 7} d u

\int \frac{2 u}{4 u ^{2} + 7} d u = \frac{1}{2} 4 u^{2} + 7

\int \frac{1}{4 u ^{2} + 7} d u = \frac{1}{2} ln \frac{2}{7} u + \frac{1}{7} (7 + 4 u^{2})

= \frac{1}{2} 4 u^{2} + 7 - \frac{1}{2} ln \frac{2}{7} u + \frac{1}{7} (7 + 4 u^{2})

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

代回

= \frac{1}{2} 4 (x + \frac{1}{2})^{2} + 7 - \frac{1}{2} ln \frac{2}{7} (x + \frac{1}{2}) + \frac{1}{7} (7 + 4 (x + \frac{1}{2})^{2})

化简

\frac{1}{2} 4 (x + \frac{1}{2})^{2} + 7 - \frac{1}{2} ln \frac{2}{7} (x + \frac{1}{2}) + \frac{1}{7} (7 + 4 (x + \frac{1}{2})^{2}) : x^{2} + x + 2 - \frac{1}{2} ln \frac{2}{7} x + \frac{1}{7} + \frac{4 x ^{2} + 4 x + 8}{7}

= x^{2} + x + 2 - \frac{1}{2} ln \frac{2}{7} x + \frac{1}{7} + \frac{4 x ^{2} + 4 x + 8}{7}

解答补常数

= x^{2} + x + 2 - \frac{1}{2} ln \frac{2}{7} x + \frac{1}{7} + \frac{4 x ^{2} + 4 x + 8}{7} + C

作图

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