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人気のある微分積分 >

integral of 4ln(2x+1)

解

\int 4 ln (2 x + 1) d x

解

2 (2 x ln (2 x + 1) + ln (2 x + 1) - 2 x - 1) + C

解答ステップ

\int 4 ln (2 x + 1) d x

定数を除く:

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

= 4 \cdot \int ln (2 x + 1) d x

u置換積分法を適用する

= 4 \cdot \int ln (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

= 4 \cdot \frac{1}{2} \cdot \int ln (u) d u

部分積分を適用する

= 4 \cdot \frac{1}{2} (u ln (u) - \int 1 d u)

\int 1 d u = u

= 4 \cdot \frac{1}{2} (u ln (u) - u)

代用を戻す

u = 2 x + 1

= 4 \cdot \frac{1}{2} ((2 x + 1) ln (2 x + 1) - (2 x + 1))

簡素化

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

= 2 (2 x ln (2 x + 1) + ln (2 x + 1) - 2 x - 1)

定数を解答に追加する

= 2 (2 x ln (2 x + 1) + ln (2 x + 1) - 2 x - 1) + C

グラフ

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