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受欢迎的微积分 >

integral of (x^2)/(sqrt(x^2+2x-3))

解答

\int \frac{x ^{2}}{x ^{2} + 2 x - 3} d x

解答

\frac{1}{2} (x + 1) x^{2} + 2 x - 3 + 3 ln (\frac{1}{2} x + x^{2} + 2 x - 3 + 1) - 2 x^{2} + 2 x - 3 + C

求解步骤

\int \frac{x ^{2}}{x ^{2} + 2 x - 3} d x

对

x^{2} + 2 x - 3

配方:

(x + 1)^{2} - 4

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

使用换元积分法

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

乘开

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

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

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

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

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

= 4 (\frac{1}{8} u u^{2} - 4 + \frac{1}{2} ln (\frac{u ^{2} - 4 + u}{2})) - 2 u^{2} - 4 + ln (\frac{1}{2} u^{2} - 4 + u)

u = x + 1

代回

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

化简

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

= \frac{1}{2} (x + 1) x^{2} + 2 x - 3 + 3 ln (\frac{1}{2} x + x^{2} + 2 x - 3 + 1) - 2 x^{2} + 2 x - 3

解答补常数

= \frac{1}{2} (x + 1) x^{2} + 2 x - 3 + 3 ln (\frac{1}{2} x + x^{2} + 2 x - 3 + 1) - 2 x^{2} + 2 x - 3 + C

作图

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