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人気のある微分積分 >

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

解

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

解

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

解答ステップ

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

三角関数による置換を適用する

= \int x^{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

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

積分換算を適用する

= x^{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) ∣

= x^{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}{x} z)

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

簡素化

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

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

定数を解答に追加する

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

グラフ

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