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人気のある微分積分 >

integral of sqrt(6+x^2)

解

\int 6 + x^{2} d x

解

6 (\frac{1}{12} x 6 + x^{2} + \frac{1}{2} ln \frac{1}{6} x + \frac{1}{6} (6 + x^{2})) + C

解答ステップ

\int 6 + x^{2} d x

三角関数による置換を適用する

= \int 6 sec^{3} (u) d u

定数を除く:

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

= 6 \cdot \int sec^{3} (u) d u

積分換算を適用する

= 6 (\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) ∣

= 6 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

代用を戻す

u = arctan (\frac{1}{6} x)

= 6 \frac{sec ^{2} ( arctan ( \frac{1}{6} x ) ) sin ( arctan ( \frac{1}{6} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{6} x)) + sec (arctan (\frac{1}{6} x))

簡素化

6 \frac{sec ^{2} ( arctan ( \frac{1}{6} x ) ) sin ( arctan ( \frac{1}{6} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{6} x)) + sec (arctan (\frac{1}{6} x)) : 6 (\frac{1}{12} x 6 + x^{2} + \frac{1}{2} ln \frac{1}{6} x + \frac{1}{6} (6 + x^{2}))

= 6 (\frac{1}{12} x 6 + x^{2} + \frac{1}{2} ln \frac{1}{6} x + \frac{1}{6} (6 + x^{2}))

定数を解答に追加する

= 6 (\frac{1}{12} x 6 + x^{2} + \frac{1}{2} ln \frac{1}{6} x + \frac{1}{6} (6 + x^{2})) + C

グラフ

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