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

解

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

ln (2) - 2 (- \frac{π}{4} + 1)

十進法表記

0.26394 \dots

解答ステップ

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

部分積分を適用する

= [x ln (1 + x^{2}) - \int \frac{2 x ^{2}}{1 + x ^{2}} d x]_{0}^{1}

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

= [x ln (1 + x^{2}) - 2 (- arctan (x) + x)]_{0}^{1}

限度を計算する:

ln (2) - 2 (- \frac{π}{4} + 1)

= ln (2) - 2 (- \frac{π}{4} + 1)