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人気のある微分積分 >

integral of 11arctan(sqrt(x))

解

\int 11 arctan (x) d x

解

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

解答ステップ

\int 11 arctan (x) d x

定数を除く:

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

= 11 \cdot \int arctan (x) d x

u置換積分法を適用する

= 11 \cdot \int arctan (u) \cdot 2 u d u

定数を除く:

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

= 11 \cdot 2 \cdot \int arctan (u) u d u

部分積分を適用する

= 11 \cdot 2 (\frac{1}{2} u^{2} arctan (u) - \int \frac{u ^{2}}{2 ( u ^{2} + 1 )} d u)

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

= 11 \cdot 2 (\frac{1}{2} u^{2} arctan (u) - \frac{1}{2} (- arctan (u) + u))

代用を戻す

u = x

= 11 \cdot 2 (\frac{1}{2} (x)^{2} arctan (x) - \frac{1}{2} (- arctan (x) + x))

簡素化

11 \cdot 2 (\frac{1}{2} (x)^{2} arctan (x) - \frac{1}{2} (- arctan (x) + x)) : 22 (\frac{1}{2} x arctan (x) - \frac{1}{2} (- arctan (x) + x))

= 22 (\frac{1}{2} x arctan (x) - \frac{1}{2} (- arctan (x) + x))

定数を解答に追加する

= 22 (\frac{1}{2} x arctan (x) - \frac{1}{2} (- arctan (x) + x)) + C

グラフ

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