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人気のある微分積分 >

integral of sqrt((x^2)/4-1)

解

\int \frac{x ^{2}}{4} - 1 d x

解

\frac{1}{4} x x^{2} - 4 - ln (\frac{1}{2} x + x^{2} - 4) + C

解答ステップ

\int \frac{x ^{2}}{4} - 1 d x

三角関数による置換を適用する

= \int 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

= 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

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

拡張

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

= 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

= 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) ∣

= 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{1}{2} x)

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

簡素化

2 (- ln tan (arcsec (\frac{1}{2} x)) + sec (arcsec (\frac{1}{2} x)) + \frac{1}{2} sec (arcsec (\frac{1}{2} x)) tan (arcsec (\frac{1}{2} x)) + \frac{1}{2} ln tan (arcsec (\frac{1}{2} x)) + sec (arcsec (\frac{1}{2} x))) : \frac{1}{4} x x^{2} - 4 - ln (\frac{1}{2} x + x^{2} - 4)

= \frac{1}{4} x x^{2} - 4 - ln (\frac{1}{2} x + x^{2} - 4)

定数を解答に追加する

= \frac{1}{4} x x^{2} - 4 - ln (\frac{1}{2} x + x^{2} - 4) + C

グラフ

人気の例

derivative of 2e^{-x}x-e^{-x}x^2

\frac{d}{d x} (2 e^{- x} x - e^{- x} x^{2})

y^'+8y=4,y(0)=3

y^{^{'}} + 8 y = 4, y (0) = 3

integral from 0 to 2 of te^{t^2}

\int_{0}^{2} t e^{t^{2}} d t

(\partial)/(\partial x)((2x-x^2)(2y-y^2))

\frac{\partial}{\partial x} ((2 x - x^{2}) (2 y - y^{2}))

limit as x approaches 1 of (2x)/(x^3+1)

x \to 1 lim (\frac{2 x}{x ^{3} + 1})