integral from-1 to 1 of ln(ln(x+3))

解

\int_{- 1}^{1} ln (ln (x + 3)) d x

ln (16 ln^{2} (2)) - Ei (ln (4)) + Ei (ln (2))

解答ステップ

\int_{- 1}^{1} ln (ln (x + 3)) d x

u置換積分法を適用する

= \int_{2}^{4} ln (ln (u)) d u

部分積分を適用する

= [u ln (ln (u)) - \int \frac{1}{ln ( u )} d u]_{2}^{4}

\int \frac{1}{ln ( u )} d u = Ei (ln (u))

= [u ln (ln (u)) - Ei (ln (u))]_{2}^{4}

限度を計算する:

4 ln (2 ln (2)) - Ei (ln (4)) - 2 ln (ln (2)) + Ei (ln (2))

= 4 ln (2 ln (2)) - Ei (ln (4)) - 2 ln (ln (2)) + Ei (ln (2))

簡素化

= ln (16 ln^{2} (2)) - Ei (ln (4)) + Ei (ln (2))