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人気のある微分積分 >

integral of xarctan(2x+3)

解

\int x arctan (2 x + 3) d x

解

\frac{1}{8} (4 x^{2} arctan (2 x + 3) - 2 x - 8 arctan (2 x + 3) + 3 ln 4 x^{2} + 12 x + 10 - 3) + C

解答ステップ

\int x arctan (2 x + 3) d x

u置換積分法を適用する

= \int \frac{arctan ( u ) ( u - 3 )}{4} d u

定数を除く:

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

= \frac{1}{4} \cdot \int arctan (u) (u - 3) d u

拡張

arctan (u) (u - 3) : u arctan (u) - 3 arctan (u)

= \frac{1}{4} \cdot \int u arctan (u) - 3 arctan (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}{4} (\int u arctan (u) d u - \int 3 arctan (u) d u)

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

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

= \frac{1}{4} (\frac{1}{2} u^{2} arctan (u) - \frac{1}{2} (- arctan (u) + u) - 3 (u arctan (u) - \frac{1}{2} ln u^{2} + 1))

代用を戻す

u = 2 x + 3

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

簡素化

\frac{1}{4} (\frac{1}{2} (2 x + 3)^{2} arctan (2 x + 3) - \frac{1}{2} (- arctan (2 x + 3) + 2 x + 3) - 3 ((2 x + 3) arctan (2 x + 3) - \frac{1}{2} ln (2 x + 3)^{2} + 1)) : \frac{1}{8} (4 x^{2} arctan (2 x + 3) - 2 x - 8 arctan (2 x + 3) + 3 ln 4 x^{2} + 12 x + 10 - 3)

= \frac{1}{8} (4 x^{2} arctan (2 x + 3) - 2 x - 8 arctan (2 x + 3) + 3 ln 4 x^{2} + 12 x + 10 - 3)

定数を解答に追加する

= \frac{1}{8} (4 x^{2} arctan (2 x + 3) - 2 x - 8 arctan (2 x + 3) + 3 ln 4 x^{2} + 12 x + 10 - 3) + C

グラフ

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