integral from 0 to 0.4 of ln(1+x^4)

解答

\int_{0}^{0.4} ln (1 + x^{4}) d x

0.00203 \dots

求解步骤

\int_{0}^{0.4} ln (1 + x^{4}) d x

使用分布积分法

= [x ln (1 + x^{4}) - \int \frac{4 x ^{4}}{1 + x ^{4}} d x]_{0}^{0.4}

\int \frac{4 x ^{4}}{1 + x ^{4}} d x = - \frac{1}{2} ln 2 x^{2} + 22 x + 2 - 2 arctan (2 x + 1) + \frac{1}{2} ln 2 x^{2} - 22 x + 2 - 2 arctan (2 x - 1) + 4 x

= [x ln (1 + x^{4}) - (- \frac{1}{2} ln 2 x^{2} + 22 x + 2 - 2 arctan (2 x + 1) + \frac{1}{2} ln 2 x^{2} - 22 x + 2 - 2 arctan (2 x - 1) + 4 x)]_{0}^{0.4}

化简

[x ln (1 + x^{4}) - (- \frac{1}{2} ln 2 x^{2} + 22 x + 2 - 2 arctan (2 x + 1) + \frac{1}{2} ln 2 x^{2} - 22 x + 2 - 2 arctan (2 x - 1) + 4 x)]_{0}^{0.4} : [x ln (1 + x^{4}) + 0.70710 \dots ln 2 x^{2} + 2.82842 \dots x + 2 + 1.41421 \dots arctan (1.41421 \dots x + 1) - 0.70710 \dots ln 2 x^{2} - 2.82842 \dots x + 2 + 1.41421 \dots arctan (1.41421 \dots x - 1) - 4 x]_{0}^{0.4}

= [x ln (1 + x^{4}) + 0.70710 \dots ln 2 x^{2} + 2.82842 \dots x + 2 + 1.41421 \dots arctan (1.41421 \dots x + 1) - 0.70710 \dots ln 2 x^{2} - 2.82842 \dots x + 2 + 1.41421 \dots arctan (1.41421 \dots x - 1) - 4 x]_{0}^{0.4}

带入上下限计算后得:

0.00203 \dots

= 0.00203 \dots

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