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人気のある微分積分 >

integral of (3x+x/((x^2+2)^{4/3)})

解

\int (3 x + \frac{x}{( x ^{2} + 2 ) ^{\frac{4}{3}}}) d x

解

\frac{3 ( x ^{2} + 2 ) ^{\frac{4}{3}} - 3}{2 ( x ^{2} + 2 ) ^{\frac{1}{3}}} + C

解答ステップ

\int 3 x + \frac{x}{( x ^{2} + 2 ) ^{\frac{4}{3}}} d x

u置換積分法を適用する

= \int \frac{3 u ^{\frac{4}{3}} + 1}{2 u ^{\frac{4}{3}}} d u

定数を除く:

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

= \frac{1}{2} \cdot \int \frac{3 u ^{\frac{4}{3}} + 1}{u ^{\frac{4}{3}}} d u

拡張

\frac{3 u ^{\frac{4}{3}} + 1}{u ^{\frac{4}{3}}} : 3 + \frac{1}{u ^{\frac{4}{3}}}

= \frac{1}{2} \cdot \int 3 + \frac{1}{u ^{\frac{4}{3}}} d u

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} (\int 3 d u + \int \frac{1}{u ^{\frac{4}{3}}} d u)

\int 3 d u = 3 u

\int \frac{1}{u ^{\frac{4}{3}}} d u = - \frac{3}{u ^{\frac{1}{3}}}

= \frac{1}{2} (3 u - \frac{3}{u ^{\frac{1}{3}}})

代用を戻す

u = x^{2} + 2

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

簡素化

\frac{1}{2} (3 (x^{2} + 2) - \frac{3}{( x ^{2} + 2 ) ^{\frac{1}{3}}}) : \frac{3 ( x ^{2} + 2 ) ^{\frac{4}{3}} - 3}{2 ( x ^{2} + 2 ) ^{\frac{1}{3}}}

= \frac{3 ( x ^{2} + 2 ) ^{\frac{4}{3}} - 3}{2 ( x ^{2} + 2 ) ^{\frac{1}{3}}}

定数を解答に追加する

= \frac{3 ( x ^{2} + 2 ) ^{\frac{4}{3}} - 3}{2 ( x ^{2} + 2 ) ^{\frac{1}{3}}} + C

グラフ

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