integral from 0 to 0.1 of ln(1+x^3)

解

\int_{0}^{0.1} ln (1 + x^{3}) d x

0.00002 \dots

解答ステップ

\int_{0}^{0.1} ln (1 + x^{3}) d x

部分積分を適用する

= [x ln (1 + x^{3}) - \int \frac{3 x ^{3}}{1 + x ^{3}} d x]_{0}^{0.1}

\int \frac{3 x ^{3}}{1 + x ^{3}} d x = - ln ∣ x + 1 ∣ - \frac{1}{2} (- ln \frac{4 x ^{2} - 4 x}{3} + \frac{4}{3} + 23 arctan (\frac{2 x - 1}{3})) + 3 x

= [x ln (1 + x^{3}) - (- ln ∣ x + 1 ∣ - \frac{1}{2} (- ln \frac{4 x ^{2} - 4 x}{3} + \frac{4}{3} + 23 arctan (\frac{2 x - 1}{3})) + 3 x)]_{0}^{0.1}

簡素化

[x ln (1 + x^{3}) - (- ln ∣ x + 1 ∣ - \frac{1}{2} (- ln \frac{4 x ^{2} - 4 x}{3} + \frac{4}{3} + 23 arctan (\frac{2 x - 1}{3})) + 3 x)]_{0}^{0.1} : [\frac{1}{2} (2 x ln (1 + x^{3}) + 2 ln ∣ x + 1 ∣ - ln \frac{4 x ^{2} - 4 x}{3} + \frac{4}{3} + 3.46410 \dots arctan (\frac{2 x - 1}{1.73205 \dots}) - 6 x)]_{0}^{0.1}

= [\frac{1}{2} (2 x ln (1 + x^{3}) + 2 ln ∣ x + 1 ∣ - ln \frac{4 x ^{2} - 4 x}{3} + \frac{4}{3} + 3.46410 \dots arctan (\frac{2 x - 1}{1.73205 \dots}) - 6 x)]_{0}^{0.1}

限度を計算する:

0.00002 \dots

= 0.00002 \dots