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人気のある微分積分 >

integral of ((x^2-4)^{5/2})/x

解

\int \frac{( x ^{2} - 4 ) ^{\frac{5}{2}}}{x} d x

解

32 (\frac{1}{160} (x^{2} - 4)^{\frac{5}{2}} - \frac{1}{24} (x^{2} - 4)^{\frac{3}{2}} + \frac{1}{2} (x^{2} - 4)^{\frac{1}{2}} - arctan (\frac{1}{2} (x^{2} - 4)^{\frac{1}{2}})) + C

解答ステップ

\int \frac{( x ^{2} - 4 ) ^{\frac{5}{2}}}{x} d x

三角関数による置換を適用する

= \int 32 tan^{6} (u) d u

定数を除く:

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

= 32 \cdot \int tan^{6} (u) d u

u置換積分法を適用する

= 32 \cdot \int \frac{v ^{6}}{1 + v ^{2}} d v

長除法

\frac{v ^{6}}{1 + v ^{2}} : v^{4} - v^{2} + 1 - \frac{1}{v ^{2} + 1}

= 32 \cdot \int v^{4} - v^{2} + 1 - \frac{1}{v ^{2} + 1} 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

= 32 (\int v^{4} d v - \int v^{2} d v + \int 1 d v - \int \frac{1}{v ^{2} + 1} d v)

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

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

\int 1 d v = v

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= 32 (\frac{v ^{5}}{5} - \frac{v ^{3}}{3} + v - arctan (v))

代用を戻す

= 32 (\frac{tan ^{5} ( arcsec ( \frac{1}{2} x ) )}{5} - \frac{tan ^{3} ( arcsec ( \frac{1}{2} x ) )}{3} + tan (arcsec (\frac{1}{2} x)) - arctan (tan (arcsec (\frac{1}{2} x))))

簡素化

32 (\frac{tan ^{5} ( arcsec ( \frac{1}{2} x ) )}{5} - \frac{tan ^{3} ( arcsec ( \frac{1}{2} x ) )}{3} + tan (arcsec (\frac{1}{2} x)) - arctan (tan (arcsec (\frac{1}{2} x)))) : 32 (\frac{1}{160} (x^{2} - 4)^{\frac{5}{2}} - \frac{1}{24} (x^{2} - 4)^{\frac{3}{2}} + \frac{1}{2} (x^{2} - 4)^{\frac{1}{2}} - arctan (\frac{1}{2} (x^{2} - 4)^{\frac{1}{2}}))

= 32 (\frac{1}{160} (x^{2} - 4)^{\frac{5}{2}} - \frac{1}{24} (x^{2} - 4)^{\frac{3}{2}} + \frac{1}{2} (x^{2} - 4)^{\frac{1}{2}} - arctan (\frac{1}{2} (x^{2} - 4)^{\frac{1}{2}}))

定数を解答に追加する

= 32 (\frac{1}{160} (x^{2} - 4)^{\frac{5}{2}} - \frac{1}{24} (x^{2} - 4)^{\frac{3}{2}} + \frac{1}{2} (x^{2} - 4)^{\frac{1}{2}} - arctan (\frac{1}{2} (x^{2} - 4)^{\frac{1}{2}})) + C

グラフ

人気の例

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