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受欢迎的微积分 >

integral of xsqrt(x^2+2x+2)

解答

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

解答

x (\frac{1}{2} (x + 1) x^{2} + 2 x + 2 + \frac{1}{2} ln x + 1 + x^{2} + 2 x + 2) - \frac{1}{6} ((x^{2} + 2 x + 2)^{\frac{3}{2}} + 3 ln x + 1 + x^{2} + 2 x + 2 (x + 1) - 3 x^{2} + 2 x + 2) + C

求解步骤

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

对

x^{2} + 2 x + 2

配方:

(x + 1)^{2} + 1

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

使用换元积分法

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

使用分布积分法

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

\int \frac{1}{2} (u 1 + u^{2} + ln u + 1 + u^{2}) d u = \frac{1}{2} (\frac{1}{3} (1 + u^{2})^{\frac{3}{2}} + u ln u + 1 + u^{2} - 1 + u^{2})

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

u = x + 1

代回

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

化简

(x + 1 - 1) (\frac{1}{2} (x + 1) 1 + (x + 1)^{2} + \frac{1}{2} ln x + 1 + 1 + (x + 1)^{2}) - \frac{1}{2} (\frac{1}{3} (1 + (x + 1)^{2})^{\frac{3}{2}} + (x + 1) ln x + 1 + 1 + (x + 1)^{2} - 1 + (x + 1)^{2}) : x (\frac{1}{2} (x + 1) x^{2} + 2 x + 2 + \frac{1}{2} ln x + 1 + x^{2} + 2 x + 2) - \frac{1}{6} ((x^{2} + 2 x + 2)^{\frac{3}{2}} + 3 ln x + 1 + x^{2} + 2 x + 2 (x + 1) - 3 x^{2} + 2 x + 2)

= x (\frac{1}{2} (x + 1) x^{2} + 2 x + 2 + \frac{1}{2} ln x + 1 + x^{2} + 2 x + 2) - \frac{1}{6} ((x^{2} + 2 x + 2)^{\frac{3}{2}} + 3 ln x + 1 + x^{2} + 2 x + 2 (x + 1) - 3 x^{2} + 2 x + 2)

解答补常数

= x (\frac{1}{2} (x + 1) x^{2} + 2 x + 2 + \frac{1}{2} ln x + 1 + x^{2} + 2 x + 2) - \frac{1}{6} ((x^{2} + 2 x + 2)^{\frac{3}{2}} + 3 ln x + 1 + x^{2} + 2 x + 2 (x + 1) - 3 x^{2} + 2 x + 2) + C

作图

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