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人気のある微分積分 >

integral of (x^5)/(sqrt(x^2+5))

解

\int \frac{x ^{5}}{x ^{2} + 5} d x

解

255 \frac{1}{5} (5 + x^{2}) - \frac{10}{3} (x^{2} + 5)^{\frac{3}{2}} + \frac{1}{5} (5 + x^{2})^{\frac{5}{2}} + C

解答ステップ

\int \frac{x ^{5}}{x ^{2} + 5} d x

三角関数による置換を適用する

= \int 255 tan^{5} (u) sec (u) d u

定数を除く:

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

= 255 \cdot \int tan^{5} (u) sec (u) d u

三角関数の公式を使用して書き換える

= 255 \cdot \int (- 1 + sec^{2} (u))^{2} tan (u) sec (u) d u

u置換積分法を適用する

= 255 \cdot \int (- 1 + v^{2})^{2} d v

拡張

(- 1 + v^{2})^{2} : 1 - 2 v^{2} + v^{4}

= 255 \cdot \int 1 - 2 v^{2} + v^{4} d v

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

= 255 (\int 1 d v - \int 2 v^{2} d v + \int v^{4} d v)

\int 1 d v = v

\int 2 v^{2} d v = \frac{2 v ^{3}}{3}

\int v^{4} d v = \frac{v ^{5}}{5}

= 255 (v - \frac{2 v ^{3}}{3} + \frac{v ^{5}}{5})

代用を戻す

= 255 sec (arctan (\frac{1}{5} x)) - \frac{2 sec ^{3} ( arctan ( \frac{1}{5} x ) )}{3} + \frac{sec ^{5} ( arctan ( \frac{1}{5} x ) )}{5}

簡素化

255 sec (arctan (\frac{1}{5} x)) - \frac{2 sec ^{3} ( arctan ( \frac{1}{5} x ) )}{3} + \frac{sec ^{5} ( arctan ( \frac{1}{5} x ) )}{5} : 255 \frac{1}{5} (5 + x^{2}) - \frac{10}{3} (x^{2} + 5)^{\frac{3}{2}} + \frac{1}{5} (5 + x^{2})^{\frac{5}{2}}

= 255 \frac{1}{5} (5 + x^{2}) - \frac{10}{3} (x^{2} + 5)^{\frac{3}{2}} + \frac{1}{5} (5 + x^{2})^{\frac{5}{2}}

定数を解答に追加する

= 255 \frac{1}{5} (5 + x^{2}) - \frac{10}{3} (x^{2} + 5)^{\frac{3}{2}} + \frac{1}{5} (5 + x^{2})^{\frac{5}{2}} + C

グラフ

人気の例

(\partial)/(\partial t)(st)

\frac{\partial}{\partial t} (s t)

integral of 3xsqrt(5+x^2)

\int 3 x 5 + x^{2} d x

integral of cos(2y)

\int cos (2 y) d y

integral of 1/(xsqrt(x))

\int \frac{1}{x x} d x

limit as x approaches 1 of 2x^2-4x+2

x \to 1 lim (2 x^{2} - 4 x + 2)