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人気のある微分積分 >

integral of sqrt(1+4t^2)

解

\int 1 + 4 t^{2} d t

解

\frac{1}{4} (2 t 4 t^{2} + 1 + ln 2 t + 1 + 4 t^{2}) + C

解答ステップ

\int 1 + 4 t^{2} d t

三角関数による置換を適用する

= \int \frac{1}{2} sec^{3} (u) d u

定数を除く:

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

= \frac{1}{2} \cdot \int sec^{3} (u) d u

積分換算を適用する

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

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

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

代用を戻す

u = arctan (2 t)

= \frac{1}{2} (\frac{sec ^{2} ( arctan ( 2 t ) ) sin ( arctan ( 2 t ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (2 t)) + sec (arctan (2 t)) ∣)

簡素化

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

= \frac{1}{4} (2 t 4 t^{2} + 1 + ln 2 t + 1 + 4 t^{2})

定数を解答に追加する

= \frac{1}{4} (2 t 4 t^{2} + 1 + ln 2 t + 1 + 4 t^{2}) + C

グラフ

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