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受欢迎的微积分 >

integral of x/(sqrt(x^2+2x-5))

解答

\int \frac{x}{x ^{2} + 2 x - 5} d x

解答

x^{2} + 2 x - 5 - ln \frac{1}{6} (x^{2} + 2 x - 5) + \frac{1}{6} (x + 1) + C

求解步骤

\int \frac{x}{x ^{2} + 2 x - 5} d x

对

x^{2} + 2 x - 5

配方:

(x + 1)^{2} - 6

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

使用换元积分法

= \int \frac{u - 1}{u ^{2} - 6} d u

乘开

\frac{u - 1}{u ^{2} - 6} : \frac{u}{u ^{2} - 6} - \frac{1}{u ^{2} - 6}

= \int \frac{u}{u ^{2} - 6} - \frac{1}{u ^{2} - 6} 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{u}{u ^{2} - 6} d u - \int \frac{1}{u ^{2} - 6} d u

\int \frac{u}{u ^{2} - 6} d u = u^{2} - 6

\int \frac{1}{u ^{2} - 6} d u = ln \frac{1}{6} (u^{2} - 6) + \frac{1}{6} u

= u^{2} - 6 - ln \frac{1}{6} (u^{2} - 6) + \frac{1}{6} u

u = x + 1

代回

= (x + 1)^{2} - 6 - ln \frac{1}{6} ((x + 1)^{2} - 6) + \frac{1}{6} (x + 1)

化简

(x + 1)^{2} - 6 - ln \frac{1}{6} ((x + 1)^{2} - 6) + \frac{1}{6} (x + 1) : x^{2} + 2 x - 5 - ln \frac{1}{6} (x^{2} + 2 x - 5) + \frac{1}{6} (x + 1)

= x^{2} + 2 x - 5 - ln \frac{1}{6} (x^{2} + 2 x - 5) + \frac{1}{6} (x + 1)

解答补常数

= x^{2} + 2 x - 5 - ln \frac{1}{6} (x^{2} + 2 x - 5) + \frac{1}{6} (x + 1) + C

作图

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