integral from 0 to x of ln(1+2t)

解

\int_{0}^{x} ln (1 + 2 t) d t

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

解答ステップ

\int_{0}^{x} ln (1 + 2 t) d t

u置換積分法を適用する

= \int_{1}^{1 + 2 x} 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

= \frac{1}{2} \cdot \int_{1}^{1 + 2 x} ln (u) d u

部分積分を適用する

= \frac{1}{2} [u ln (u) - \int 1 d u]_{1}^{1 + 2 x}

\int 1 d u = u

= \frac{1}{2} [u ln (u) - u]_{1}^{1 + 2 x}

限度を計算する:

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

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

簡素化

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