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人気のある微分積分 >

integral of x^5sqrt(1+x^2)

解

\int x^{5} 1 + x^{2} d x

解

\frac{1}{7} (1 + x^{2})^{\frac{7}{2}} - \frac{2}{5} (1 + x^{2})^{\frac{5}{2}} + \frac{1}{3} (1 + x^{2})^{\frac{3}{2}} + C

解答ステップ

\int x^{5} 1 + x^{2} d x

三角関数による置換を適用する

= \int tan^{5} (u) sec^{3} (u) d u

三角関数の公式を使用して書き換える

= \int (- 1 + sec^{2} (u))^{2} tan (u) sec^{3} (u) d u

u置換積分法を適用する

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

拡張

v^{2} (v^{2} - 1)^{2} : v^{6} - 2 v^{4} + v^{2}

= \int v^{6} - 2 v^{4} + v^{2} 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

= \int v^{6} d v - \int 2 v^{4} d v + \int v^{2} d v

\int v^{6} d v = \frac{v ^{7}}{7}

\int 2 v^{4} d v = \frac{2 v ^{5}}{5}

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

= \frac{v ^{7}}{7} - \frac{2 v ^{5}}{5} + \frac{v ^{3}}{3}

代用を戻す

= \frac{sec ^{7} ( arctan ( x ) )}{7} - \frac{2 sec ^{5} ( arctan ( x ) )}{5} + \frac{sec ^{3} ( arctan ( x ) )}{3}

簡素化

\frac{sec ^{7} ( arctan ( x ) )}{7} - \frac{2 sec ^{5} ( arctan ( x ) )}{5} + \frac{sec ^{3} ( arctan ( x ) )}{3} : \frac{1}{7} (1 + x^{2})^{\frac{7}{2}} - \frac{2}{5} (1 + x^{2})^{\frac{5}{2}} + \frac{1}{3} (1 + x^{2})^{\frac{3}{2}}

= \frac{1}{7} (1 + x^{2})^{\frac{7}{2}} - \frac{2}{5} (1 + x^{2})^{\frac{5}{2}} + \frac{1}{3} (1 + x^{2})^{\frac{3}{2}}

定数を解答に追加する

= \frac{1}{7} (1 + x^{2})^{\frac{7}{2}} - \frac{2}{5} (1 + x^{2})^{\frac{5}{2}} + \frac{1}{3} (1 + x^{2})^{\frac{3}{2}} + C

グラフ

人気の例

y^{^{''}} = k y

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