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人気のある微分積分 >

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

解

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

解

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

解答ステップ

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

平方完成する

x^{2} + 2 x + 2 : (x + 1)^{2} + 1

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

u置換積分法を適用する

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

部分積分を適用する

= (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})

代用を戻す

u = x + 1

= (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})

簡素化

(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)

定数を解答に追加する

= 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

グラフ

人気の例

integral of xsin(-2x)

\int x sin (- 2 x) d x

9y^{''}+24y^'+16y=0

9 y^{^{''}} + 24 y^{^{'}} + 16 y = 0

(\frac{1}{1 + x ^{2}})^{^{'}}

integral from-1 to 5 of 3x^2+6x

\int_{- 1}^{5} 3 x^{2} + 6 x d x

tangent y=4x^3-4x,(-1,0)

t an g e n t y = 4 x^{3} - 4 x, (- 1, 0)