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人気のある微分積分 >

integral of sqrt(x(x+1))

解

\int x (x + 1) d x

解

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

解答ステップ

\int x (x + 1) d x

u置換積分法を適用する

= \int (u - 1) u d u

拡張

(u - 1) u : u^{2} - u

= \int u^{2} - u d u

平方完成する

u^{2} - u : (u - \frac{1}{2})^{2} - \frac{1}{4}

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

u置換積分法を適用する

= \int \frac{4 v ^{2} - 1}{2} d v

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int 4 v^{2} - 1 d v

三角関数による置換を適用する

= \frac{1}{2} \cdot \int \frac{1}{2} tan^{2} (w) sec (w) d w

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int tan^{2} (w) sec (w) d w

三角関数の公式を使用して書き換える

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

拡張

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

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

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

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

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

= \frac{1}{2} \cdot \frac{1}{2} (- ln ∣ tan (w) + sec (w) ∣ + \frac{1}{2} sec (w) tan (w) + \frac{1}{2} ln ∣ tan (w) + sec (w) ∣)

代用を戻す

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

簡素化

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

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

定数を解答に追加する

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

グラフ

人気の例

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\int_{1}^{100} \frac{1}{x ^{2}} d x

implicit (dy)/(dx),2x=y^4

im pl i c i t \frac{d y}{d x}, 2 x = y^{4}

limit as x approaches 0 of x/(-4x^3-8x)

x \to 0 lim (\frac{x}{- 4 x ^{3} - 8 x})

(\partial}{\partial t}(\frac{(s-3r))/t)

\frac{\partial}{\partial t} (\frac{( s - 3 r )}{t})

integral of 2x(x^2+1)^4

\int 2 x (x^{2} + 1)^{4} d x