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人気のある微分積分 >

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

解

\int arctan (2 x + 1) d x

解

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

解答ステップ

\int arctan (2 x + 1) d x

u置換積分法を適用する

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

u置換積分法を適用する

= \frac{1}{2} \cdot \int arctan (v) \cdot 2 v d v

定数を除く:

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

= \frac{1}{2} \cdot 2 \cdot \int arctan (v) v d v

部分積分を適用する

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

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

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

代用を戻す

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

簡素化

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

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

定数を解答に追加する

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

グラフ

人気の例

integral of csc^2(x)x

\int csc^{2} (x) x d x

integral of x^3(x^2+1)^{12}

\int x^{3} (x^{2} + 1)^{12} d x

integral of (x^{3/2}+7)^8sqrt(x)

\int (x^{\frac{3}{2}} + 7)^{8} x d x

integral of 1/(t^2sqrt(361-t^2))

\int \frac{1}{t ^{2} 361 - t ^{2}} d t

integral of y^2(7+y)^2

\int y^{2} (7 + y)^{2} d y