ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある微分積分 >

integral of sqrt(x^2-25)

解

\int x^{2} - 25 d x

解

25 (\frac{1}{50} x x^{2} - 25 - \frac{1}{2} ln (\frac{x + x ^{2} - 25}{5})) + C

解答ステップ

\int x^{2} - 25 d x

三角関数による置換を適用する

= \int 25 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

= 25 \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

= 25 \cdot \int (- 1 + sec^{2} (u)) sec (u) d u

拡張

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

= 25 \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

= 25 (- \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) ∣

= 25 (- 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}{5} x)

= 25 (- ln tan (arcsec (\frac{1}{5} x)) + sec (arcsec (\frac{1}{5} x)) + \frac{1}{2} sec (arcsec (\frac{1}{5} x)) tan (arcsec (\frac{1}{5} x)) + \frac{1}{2} ln tan (arcsec (\frac{1}{5} x)) + sec (arcsec (\frac{1}{5} x)))

簡素化

25 (- ln tan (arcsec (\frac{1}{5} x)) + sec (arcsec (\frac{1}{5} x)) + \frac{1}{2} sec (arcsec (\frac{1}{5} x)) tan (arcsec (\frac{1}{5} x)) + \frac{1}{2} ln tan (arcsec (\frac{1}{5} x)) + sec (arcsec (\frac{1}{5} x))) : 25 (\frac{1}{50} x x^{2} - 25 - \frac{1}{2} ln (\frac{x + x ^{2} - 25}{5}))

= 25 (\frac{1}{50} x x^{2} - 25 - \frac{1}{2} ln (\frac{x + x ^{2} - 25}{5}))

定数を解答に追加する

= 25 (\frac{1}{50} x x^{2} - 25 - \frac{1}{2} ln (\frac{x + x ^{2} - 25}{5})) + C

グラフ

人気の例

derivative f(x)=(x+1/(\sqrt[3]{x)})^2

d er i v a t i v e f (x) = (x + \frac{1}{3 x})^{2}

limit as x approaches infinity of e^{-6x}cos(x)

x \to \infty lim (e^{- 6 x} cos (x))

xy^2y^'=y^3-3x^3,y(1)=6

x y^{2} y^{^{'}} = y^{3} - 3 x^{3}, y (1) = 6

面積 x=8y^2,x=2+6y^2

a re a x = 8 y^{2}, x = 2 + 6 y^{2}

limit as x approaches-infinity of 10^x

x \to - \infty lim (1 0^{x})