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人気のある微分積分 >

integral of sqrt(16x^2-1)

解

\int 16 x^{2} - 1 d x

解

\frac{1}{8} (4 x 16 x^{2} - 1 - ln 4 x + 16 x^{2} - 1) + C

解答ステップ

\int 16 x^{2} - 1 d x

三角関数による置換を適用する

= \int \frac{1}{4} 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{1}{4} \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{1}{4} \cdot \int (- 1 + sec^{2} (u)) sec (u) d u

拡張

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

= \frac{1}{4} \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{1}{4} (- \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{1}{4} (- 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 (4 x)

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

簡素化

\frac{1}{4} (- ln ∣ tan (arcsec (4 x)) + sec (arcsec (4 x)) ∣ + \frac{1}{2} sec (arcsec (4 x)) tan (arcsec (4 x)) + \frac{1}{2} ln ∣ tan (arcsec (4 x)) + sec (arcsec (4 x)) ∣) : \frac{1}{8} (4 x 16 x^{2} - 1 - ln 4 x + 16 x^{2} - 1)

= \frac{1}{8} (4 x 16 x^{2} - 1 - ln 4 x + 16 x^{2} - 1)

定数を解答に追加する

= \frac{1}{8} (4 x 16 x^{2} - 1 - ln 4 x + 16 x^{2} - 1) + C

グラフ

人気の例

integral of 1/(vln(v))

\int \frac{1}{v ln ( v )} d v

面積 y=3sin(x),y=e^{7x},x=0,x= pi/2

a re a y = 3 sin (x), y = e^{7 x}, x = 0, x = \frac{π}{2}

(\partial)/(\partial x)(y^5sin(3x))

\frac{\partial}{\partial x} (y^{5} sin (3 x))

integral from 0 to 2 of 6^x

\int_{0}^{2} 6^{x} d x

limit as x approaches 0 of 9/x-9/(x^2+x)

x \to 0 lim (\frac{9}{x} - \frac{9}{x ^{2} + x})