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人気のある微分積分 >

integral from 0 to 2 of ln(1+x^4)

解

\int_{0}^{2} ln (1 + x^{4}) d x

解

2 ln (17) + 2 arctan (22 + 1) + 2 arctan (22 - 1) + \frac{ln ( \frac{33 + 20 2}{17} )}{2} - 8

+1

十進法表記

1.94693 \dots

解答ステップ

\int_{0}^{2} ln (1 + x^{4}) d x

部分積分を適用する

= [x ln (1 + x^{4}) - \int \frac{4 x ^{4}}{1 + x ^{4}} d x]_{0}^{2}

\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}^{2}

簡素化

= [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}^{2}

限度を計算する:

2 ln (17) + 2 arctan (22 + 1) + 2 arctan (22 - 1) + \frac{ln ( 10 + 4 2 ) - ln ( 10 - 4 2 )}{2} - 8

= 2 ln (17) + 2 arctan (22 + 1) + 2 arctan (22 - 1) + \frac{ln ( 10 + 4 2 ) - ln ( 10 - 4 2 )}{2} - 8

簡素化

= 2 ln (17) + 2 arctan (22 + 1) + 2 arctan (22 - 1) + \frac{ln ( \frac{33 + 20 2}{17} )}{2} - 8

グラフ

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