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人気のある微分積分 >

integral of x/(2(x+1)sqrt(2x+1))

解

\int \frac{x}{2 ( x + 1 ) 2 x + 1} d x

解

\frac{1}{2} (- 2 arctan (2 x + 1) + 2 x + 1) + C

解答ステップ

\int \frac{x}{2 ( x + 1 ) 2 x + 1} d x

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{x}{( x + 1 ) 2 x + 1} d x

u置換積分法を適用する

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

拡張

\frac{u - 1}{u ( u + 1 )} : \frac{u}{u ( u + 1 )} - \frac{1}{u ( u + 1 )}

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} \cdot \frac{1}{2} (\int \frac{u}{u ( u + 1 )} d u - \int \frac{1}{u ( u + 1 )} d u)

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

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

= \frac{1}{2} \cdot \frac{1}{2} (2 (- arctan (u) + u) - 2 arctan (u))

代用を戻す

u = 2 x + 1

= \frac{1}{2} \cdot \frac{1}{2} (2 (- arctan (2 x + 1) + 2 x + 1) - 2 arctan (2 x + 1))

簡素化

\frac{1}{2} \cdot \frac{1}{2} (2 (- arctan (2 x + 1) + 2 x + 1) - 2 arctan (2 x + 1)) : \frac{1}{2} (- 2 arctan (2 x + 1) + 2 x + 1)

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

定数を解答に追加する

= \frac{1}{2} (- 2 arctan (2 x + 1) + 2 x + 1) + C

グラフ

人気の例

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