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人気のある微分積分 >

integral of sqrt(2x^2-3)

解

\int 2 x^{2} - 3 d x

解

\frac{3}{2} (\frac{1}{6} x \frac{1}{3} (2 x^{2} - 3) - \frac{1}{2} ln \frac{2}{3} x + \frac{1}{3} (2 x^{2} - 3)) + C

解答ステップ

\int 2 x^{2} - 3 d x

三角関数による置換を適用する

= \int \frac{3}{2} 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

= \frac{3}{2} \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

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

拡張

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

= \frac{3}{2} \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

= \frac{3}{2} (- \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) ∣

= \frac{3}{2} (- 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{2}{3} x)

= \frac{3}{2} (- ln tan (arcsec (\frac{2}{3} x)) + sec (arcsec (\frac{2}{3} x)) + \frac{1}{2} sec (arcsec (\frac{2}{3} x)) tan (arcsec (\frac{2}{3} x)) + \frac{1}{2} ln tan (arcsec (\frac{2}{3} x)) + sec (arcsec (\frac{2}{3} x)))

簡素化

\frac{3}{2} (- ln tan (arcsec (\frac{2}{3} x)) + sec (arcsec (\frac{2}{3} x)) + \frac{1}{2} sec (arcsec (\frac{2}{3} x)) tan (arcsec (\frac{2}{3} x)) + \frac{1}{2} ln tan (arcsec (\frac{2}{3} x)) + sec (arcsec (\frac{2}{3} x))) : \frac{3}{2} (\frac{1}{6} x \frac{1}{3} (2 x^{2} - 3) - \frac{1}{2} ln \frac{2}{3} x + \frac{1}{3} (2 x^{2} - 3))

= \frac{3}{2} (\frac{1}{6} x \frac{1}{3} (2 x^{2} - 3) - \frac{1}{2} ln \frac{2}{3} x + \frac{1}{3} (2 x^{2} - 3))

定数を解答に追加する

= \frac{3}{2} (\frac{1}{6} x \frac{1}{3} (2 x^{2} - 3) - \frac{1}{2} ln \frac{2}{3} x + \frac{1}{3} (2 x^{2} - 3)) + C

グラフ

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