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

解

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

x ln (x^{2} + 1) - 2 (x - arctan (x))

解答ステップ

\int_{0}^{x} 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}^{x}

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

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

限度を計算する:

x ln (x^{2} + 1) - 2 (x - arctan (x))

= x ln (x^{2} + 1) - 2 (x - arctan (x))