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人気のある微分積分 >

integral of x^3(1+2x^2)^{(-3)/2}

解

\int x^{3} (1 + 2 x^{2})^{\frac{- 3}{2}} d x

解

\frac{x ^{2} + 1}{2 ( 2 x ^{2} + 1 ) ^{\frac{1}{2}}} + C

解答ステップ

\int x^{3} (1 + 2 x^{2})^{\frac{- 3}{2}} d x

\frac{- 3}{2} = - \frac{3}{2}

= \int x^{3} (1 + 2 x^{2})^{- \frac{3}{2}} d x

u置換積分法を適用する

= \int \frac{u ^{\frac{1}{3}}}{3 ( 2 u ^{\frac{2}{3}} + 1 ) ^{\frac{3}{2}}} d u

定数を除く:

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

= \frac{1}{3} \cdot \int \frac{u ^{\frac{1}{3}}}{( 2 u ^{\frac{2}{3}} + 1 ) ^{\frac{3}{2}}} d u

u置換積分法を適用する

= \frac{1}{3} \cdot \int \frac{3 ( v - 1 )}{8 v ^{\frac{3}{2}}} d v

定数を除く:

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

= \frac{1}{3} \cdot \frac{3}{8} \cdot \int \frac{v - 1}{v ^{\frac{3}{2}}} d v

拡張

\frac{v - 1}{v ^{\frac{3}{2}}} : \frac{1}{v ^{\frac{1}{2}}} - \frac{1}{v ^{\frac{3}{2}}}

= \frac{1}{3} \cdot \frac{3}{8} \cdot \int \frac{1}{v ^{\frac{1}{2}}} - \frac{1}{v ^{\frac{3}{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

= \frac{1}{3} \cdot \frac{3}{8} (\int \frac{1}{v ^{\frac{1}{2}}} d v - \int \frac{1}{v ^{\frac{3}{2}}} d v)

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

\int \frac{1}{v ^{\frac{3}{2}}} d v = - \frac{2}{v ^{\frac{1}{2}}}

= \frac{1}{3} \cdot \frac{3}{8} (2 v^{\frac{1}{2}} - (- \frac{2}{v ^{\frac{1}{2}}}))

代用を戻す

= \frac{1}{3} \cdot \frac{3}{8} 2 (2 (x^{3})^{\frac{2}{3}} + 1)^{\frac{1}{2}} - - \frac{2}{( 2 ( x ^{3} ) ^{\frac{2}{3}} + 1 ) ^{\frac{1}{2}}}

簡素化

\frac{1}{3} \cdot \frac{3}{8} 2 (2 (x^{3})^{\frac{2}{3}} + 1)^{\frac{1}{2}} - - \frac{2}{( 2 ( x ^{3} ) ^{\frac{2}{3}} + 1 ) ^{\frac{1}{2}}} : \frac{x ^{2} + 1}{2 ( 2 x ^{2} + 1 ) ^{\frac{1}{2}}}

= \frac{x ^{2} + 1}{2 ( 2 x ^{2} + 1 ) ^{\frac{1}{2}}}

定数を解答に追加する

= \frac{x ^{2} + 1}{2 ( 2 x ^{2} + 1 ) ^{\frac{1}{2}}} + C

グラフ

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