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人気のある微分積分 >

integral of 2x^7sqrt(x^4+1)

解

\int 2 x^{7} x^{4} + 1 d x

解

\frac{1}{15} (3 x^{4} - 2) (x^{4} + 1)^{\frac{3}{2}} + C

解答ステップ

\int 2 x^{7} x^{4} + 1 d x

定数を除く:

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

= 2 \cdot \int x^{7} x^{4} + 1 d x

u置換積分法を適用する

= 2 \cdot \int \frac{u ^{\frac{1}{7}} u ^{\frac{4}{7}} + 1}{7} d u

定数を除く:

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

= 2 \cdot \frac{1}{7} \cdot \int u^{\frac{1}{7}} u^{\frac{4}{7}} + 1 d u

u置換積分法を適用する

= 2 \cdot \frac{1}{7} \cdot \int \frac{7 ( v - 1 ) v}{4} d v

定数を除く:

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

= 2 \cdot \frac{1}{7} \cdot \frac{7}{4} \cdot \int (v - 1) v d v

拡張

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

= 2 \cdot \frac{1}{7} \cdot \frac{7}{4} \cdot \int v^{\frac{3}{2}} - v 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

= 2 \cdot \frac{1}{7} \cdot \frac{7}{4} (\int v^{\frac{3}{2}} d v - \int v d v)

\int v^{\frac{3}{2}} d v = \frac{2}{5} v^{\frac{5}{2}}

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

= 2 \cdot \frac{1}{7} \cdot \frac{7}{4} (\frac{2}{5} v^{\frac{5}{2}} - \frac{2}{3} v^{\frac{3}{2}})

代用を戻す

= 2 \cdot \frac{1}{7} \cdot \frac{7}{4} (\frac{2}{5} ((x^{7})^{\frac{4}{7}} + 1)^{\frac{5}{2}} - \frac{2}{3} ((x^{7})^{\frac{4}{7}} + 1)^{\frac{3}{2}})

簡素化

2 \cdot \frac{1}{7} \cdot \frac{7}{4} (\frac{2}{5} ((x^{7})^{\frac{4}{7}} + 1)^{\frac{5}{2}} - \frac{2}{3} ((x^{7})^{\frac{4}{7}} + 1)^{\frac{3}{2}}) : \frac{1}{15} (3 x^{4} - 2) (x^{4} + 1)^{\frac{3}{2}}

= \frac{1}{15} (3 x^{4} - 2) (x^{4} + 1)^{\frac{3}{2}}

定数を解答に追加する

= \frac{1}{15} (3 x^{4} - 2) (x^{4} + 1)^{\frac{3}{2}} + C

グラフ

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