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人気のある微分積分 >

integral of (x^3+5x^2-2)e^{2x}

解

\int (x^{3} + 5 x^{2} - 2) e^{2 x} d x

解

\frac{1}{16} (8 e^{2 x} x^{3} - 3 (4 e^{2 x} x^{2} - 2 (2 e^{2 x} x - e^{2 x}))) + \frac{5}{4} e^{2 x} (2 x^{2} - 2 x + 1) - e^{2 x} + C

解答ステップ

\int (x^{3} + 5 x^{2} - 2) e^{2 x} d x

u置換積分法を適用する

= \int \frac{e ^{u} u ^{3}}{16} + \frac{5 e ^{u} u ^{2}}{8} - e^{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

= \int \frac{e ^{u} u ^{3}}{16} d u + \int \frac{5 e ^{u} u ^{2}}{8} d u - \int e^{u} d u

\int \frac{e ^{u} u ^{3}}{16} d u = \frac{1}{16} (u^{3} e^{u} - 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u})))

\int \frac{5 e ^{u} u ^{2}}{8} d u = \frac{5}{8} (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

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

= \frac{1}{16} (u^{3} e^{u} - 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u}))) + \frac{5}{8} (u^{2} e^{u} - 2 (e^{u} u - e^{u})) - e^{u}

代用を戻す

u = 2 x

= \frac{1}{16} ((2 x)^{3} e^{2 x} - 3 ((2 x)^{2} e^{2 x} - 2 (e^{2 x} \cdot 2 x - e^{2 x}))) + \frac{5}{8} ((2 x)^{2} e^{2 x} - 2 (e^{2 x} \cdot 2 x - e^{2 x})) - e^{2 x}

簡素化

\frac{1}{16} ((2 x)^{3} e^{2 x} - 3 ((2 x)^{2} e^{2 x} - 2 (e^{2 x} \cdot 2 x - e^{2 x}))) + \frac{5}{8} ((2 x)^{2} e^{2 x} - 2 (e^{2 x} \cdot 2 x - e^{2 x})) - e^{2 x} : \frac{1}{16} (8 e^{2 x} x^{3} - 3 (4 e^{2 x} x^{2} - 2 (2 e^{2 x} x - e^{2 x}))) + \frac{5}{4} e^{2 x} (2 x^{2} - 2 x + 1) - e^{2 x}

= \frac{1}{16} (8 e^{2 x} x^{3} - 3 (4 e^{2 x} x^{2} - 2 (2 e^{2 x} x - e^{2 x}))) + \frac{5}{4} e^{2 x} (2 x^{2} - 2 x + 1) - e^{2 x}

定数を解答に追加する

= \frac{1}{16} (8 e^{2 x} x^{3} - 3 (4 e^{2 x} x^{2} - 2 (2 e^{2 x} x - e^{2 x}))) + \frac{5}{4} e^{2 x} (2 x^{2} - 2 x + 1) - e^{2 x} + C

グラフ

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