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人気のある微分積分 >

integral of sqrt(x^2+a^2)

解

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

解

\frac{1}{2} x x^{2} + a^{2} + \frac{a ^{2}}{2} ln (\frac{x + a ^{2} + x ^{2}}{∣ a ∣}) + C

解答ステップ

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

三角関数による置換を適用する

= \int a^{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

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

積分換算を適用する

= a^{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) ∣

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

代用を戻す

u = arctan (\frac{1}{a} x)

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

簡素化

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

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

定数を解答に追加する

= \frac{1}{2} x x^{2} + a^{2} + \frac{a ^{2}}{2} ln (\frac{x + a ^{2} + x ^{2}}{∣ a ∣}) + C

グラフ

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