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受欢迎的微积分 >

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

解答

\int \frac{x}{x ^{2} + x - 1} d x

解答

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

求解步骤

\int \frac{x}{x ^{2} + x - 1} d x

对

x^{2} + x - 1

配方:

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

= \int \frac{x}{( x + \frac{1}{2} ) ^{2} - \frac{5}{4}} d x

使用换元积分法

= \int \frac{2 ( 2 u - 1 )}{4 u ^{2} - 5} d u

提出常数:

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

= 2 \cdot \int \frac{2 u - 1}{4 u ^{2} - 5} d u

乘开

\frac{2 u - 1}{4 u ^{2} - 5} : \frac{2 u}{4 u ^{2} - 5} - \frac{1}{4 u ^{2} - 5}

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 (\int \frac{2 u}{4 u ^{2} - 5} d u - \int \frac{1}{4 u ^{2} - 5} d u)

\int \frac{2 u}{4 u ^{2} - 5} d u = \frac{1}{4} ln 4 u^{2} - 5

\int \frac{1}{4 u ^{2} - 5} d u = - \frac{1}{4 5} (ln \frac{2}{5} u + 1 - ln \frac{2}{5} u - 1)

= 2 (\frac{1}{4} ln 4 u^{2} - 5 - (- \frac{1}{4 5} (ln \frac{2}{5} u + 1 - ln \frac{2}{5} u - 1)))

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

代回

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

化简

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

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

解答补常数

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

作图

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