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受欢迎的微积分 >

integral of (x^2)/(sqrt(x^2+4x-12))

解答

\int \frac{x ^{2}}{x ^{2} + 4 x - 12} d x

解答

\frac{1}{2} (x + 2) x^{2} + 4 x - 12 + 12 ln (\frac{1}{4} x + x^{2} + 4 x - 12 + 2) - 4 x^{2} + 4 x - 12 + C

求解步骤

\int \frac{x ^{2}}{x ^{2} + 4 x - 12} d x

分解

\frac{x ^{2}}{x ^{2} + 4 x - 12}

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

使用换元积分法

= \int \frac{( u - 6 ) ^{2}}{u ( u - 8 )} d u

乘开

\frac{( u - 6 ) ^{2}}{u ( u - 8 )} : \frac{u ^{2}}{u ^{2} - 8 u} - \frac{12 u}{u ^{2} - 8 u} + \frac{36}{u ^{2} - 8 u}

= \int \frac{u ^{2}}{u ^{2} - 8 u} - \frac{12 u}{u ^{2} - 8 u} + \frac{36}{u ^{2} - 8 u} 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} - 8 u} d u - \int \frac{12 u}{u ^{2} - 8 u} d u + \int \frac{36}{u ^{2} - 8 u} d u

\int \frac{u ^{2}}{u ^{2} - 8 u} d u = \frac{1}{2} (u - 4) u^{2} - 8 u + 24 ln (\frac{1}{4} u + u^{2} - 8 u - 4) + 8 u^{2} - 8 u

\int \frac{12 u}{u ^{2} - 8 u} d u = 12 (u^{2} - 8 u + 4 ln (\frac{1}{4} u^{2} - 8 u + u - 4))

\int \frac{36}{u ^{2} - 8 u} d u = 36 ln (\frac{1}{4} u^{2} - 8 u + u - 4)

= \frac{1}{2} (u - 4) u^{2} - 8 u + 24 ln (\frac{1}{4} u + u^{2} - 8 u - 4) + 8 u^{2} - 8 u - 12 (u^{2} - 8 u + 4 ln (\frac{1}{4} u^{2} - 8 u + u - 4)) + 36 ln (\frac{1}{4} u^{2} - 8 u + u - 4)

u = x + 6

代回

= \frac{1}{2} (x + 6 - 4) (x + 6)^{2} - 8 (x + 6) + 24 ln (\frac{1}{4} x + 6 + (x + 6)^{2} - 8 (x + 6) - 4) + 8 (x + 6)^{2} - 8 (x + 6) - 12 ((x + 6)^{2} - 8 (x + 6) + 4 ln (\frac{1}{4} (x + 6)^{2} - 8 (x + 6) + x + 6 - 4)) + 36 ln (\frac{1}{4} (x + 6)^{2} - 8 (x + 6) + x + 6 - 4)

化简

\frac{1}{2} (x + 6 - 4) (x + 6)^{2} - 8 (x + 6) + 24 ln (\frac{1}{4} x + 6 + (x + 6)^{2} - 8 (x + 6) - 4) + 8 (x + 6)^{2} - 8 (x + 6) - 12 ((x + 6)^{2} - 8 (x + 6) + 4 ln (\frac{1}{4} (x + 6)^{2} - 8 (x + 6) + x + 6 - 4)) + 36 ln (\frac{1}{4} (x + 6)^{2} - 8 (x + 6) + x + 6 - 4) : \frac{1}{2} (x + 2) x^{2} + 4 x - 12 + 12 ln (\frac{1}{4} x + x^{2} + 4 x - 12 + 2) - 4 x^{2} + 4 x - 12

= \frac{1}{2} (x + 2) x^{2} + 4 x - 12 + 12 ln (\frac{1}{4} x + x^{2} + 4 x - 12 + 2) - 4 x^{2} + 4 x - 12

解答补常数

= \frac{1}{2} (x + 2) x^{2} + 4 x - 12 + 12 ln (\frac{1}{4} x + x^{2} + 4 x - 12 + 2) - 4 x^{2} + 4 x - 12 + C

作图

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