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人気のある微分積分 >

積分記号 (u^2)(1+u^2)^{0.5}

解

積分記号

(u^{2}) (1 + u^{2})^{0.5}

解

- \frac{1}{4} (\frac{1}{2} u 1 + u^{2} + \frac{1}{2} ln u + 1 + u^{2}) + \frac{1}{4} u (u^{2} + 1)^{\frac{3}{2}} + C

解答ステップ

\int u^{2} (1 + u^{2})^{0.5} d u

三角関数による置換を適用する

= \int sec^{3} (v) tan^{2} (v) d v

三角関数の公式を使用して書き換える

= \int sec^{3} (v) (- 1 + sec^{2} (v)) d v

拡張

sec^{3} (v) (- 1 + sec^{2} (v)) : - sec^{3} (v) + sec^{5} (v)

= \int - sec^{3} (v) + sec^{5} (v) d v

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

= - \int sec^{3} (v) d v + \int sec^{5} (v) d v

\int sec^{3} (v) d v = \frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

\int sec^{5} (v) d v = \frac{sec ^{4} ( v ) sin ( v )}{4} + \frac{3}{4} (\frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

= - (\frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣) + \frac{sec ^{4} ( v ) sin ( v )}{4} + \frac{3}{4} (\frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

代用を戻す

v = arctan (u)

= - (\frac{1}{2} sec (arctan (u)) tan (arctan (u)) + \frac{1}{2} ln ∣ tan (arctan (u)) + sec (arctan (u)) ∣) + \frac{sec ^{4} ( arctan ( u ) ) sin ( arctan ( u ) )}{4} + \frac{3}{4} (\frac{1}{2} sec (arctan (u)) tan (arctan (u)) + \frac{1}{2} ln ∣ tan (arctan (u)) + sec (arctan (u)) ∣)

簡素化

- (\frac{1}{2} sec (arctan (u)) tan (arctan (u)) + \frac{1}{2} ln ∣ tan (arctan (u)) + sec (arctan (u)) ∣) + \frac{sec ^{4} ( arctan ( u ) ) sin ( arctan ( u ) )}{4} + \frac{3}{4} (\frac{1}{2} sec (arctan (u)) tan (arctan (u)) + \frac{1}{2} ln ∣ tan (arctan (u)) + sec (arctan (u)) ∣) : - \frac{1}{4} (\frac{1}{2} u 1 + u^{2} + \frac{1}{2} ln u + 1 + u^{2}) + \frac{1}{4} u (u^{2} + 1)^{\frac{3}{2}}

= - \frac{1}{4} (\frac{1}{2} u 1 + u^{2} + \frac{1}{2} ln u + 1 + u^{2}) + \frac{1}{4} u (u^{2} + 1)^{\frac{3}{2}}

定数を解答に追加する

= - \frac{1}{4} (\frac{1}{2} u 1 + u^{2} + \frac{1}{2} ln u + 1 + u^{2}) + \frac{1}{4} u (u^{2} + 1)^{\frac{3}{2}} + C

グラフ

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