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人気のある微分積分 >

integral of e^{2x}(x^2+2x)

解

\int e^{2 x} (x^{2} + 2 x) d x

解

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

解答ステップ

\int e^{2 x} (x^{2} + 2 x) d x

u置換積分法を適用する

= \int \frac{1}{2} e^{u} (\frac{u ^{2}}{4} + u) 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 e^{u} (\frac{u ^{2}}{4} + u) d u

拡張

e^{u} (\frac{u ^{2}}{4} + u) : \frac{e ^{u} u ^{2}}{4} + e^{u} u

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

\int \frac{e ^{u} u ^{2}}{4} d u = \frac{1}{4} (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

\int e^{u} u d u = e^{u} u - e^{u}

= \frac{1}{2} (\frac{1}{4} (u^{2} e^{u} - 2 (e^{u} u - e^{u})) + e^{u} u - e^{u})

代用を戻す

u = 2 x

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

簡素化

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

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

定数を解答に追加する

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

グラフ

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