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人気のある微分積分 >

integral of sqrt(1+(x(x^2+2)^{1/2))^2}

解

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

解

\frac{x ^{3}}{3} + x + C

解答ステップ

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

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

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

置換積分法を適用する

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

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

(2 u^{2} + 2) = 2 (u^{2} + 1)

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

簡素化

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

拡張

4 u^{2} (u^{2} + 1) : 4 u^{4} + 4 u^{2}

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

因数

4 u^{4} + 4 u^{2} + 1 : (2 u^{2} + 1)^{2}

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

(2 u^{2} + 1)^{2} = ((2 u^{2} + 1)), （ (2 u^{2} + 1) \geq 0

と仮定

）

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

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 2 u^{2} d u = \frac{2 u ^{3}}{3}

\int 1 d u = u

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

代用を戻す

u = \frac{x}{2 ^{\frac{1}{2}}}

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

簡素化

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

= \frac{x ^{3}}{3} + x

定数を解答に追加する

= \frac{x ^{3}}{3} + x + C

グラフ

人気の例

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