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人気のある微分積分 >

integral of sqrt(x^2-6)

解

\int x^{2} - 6 d x

解

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

解答ステップ

\int x^{2} - 6 d x

三角関数による置換を適用する

= \int 6 tan^{2} (u) sec (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 tan^{2} (u) sec (u) d u

三角関数の公式を使用して書き換える

= 6 \cdot \int (- 1 + sec^{2} (u)) sec (u) d u

拡張

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

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

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

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

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

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

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

代用を戻す

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

= 6 (- ln tan (arcsec (\frac{1}{6} x)) + sec (arcsec (\frac{1}{6} x)) + \frac{1}{2} sec (arcsec (\frac{1}{6} x)) tan (arcsec (\frac{1}{6} x)) + \frac{1}{2} ln tan (arcsec (\frac{1}{6} x)) + sec (arcsec (\frac{1}{6} x)))

簡素化

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

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

定数を解答に追加する

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

グラフ

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