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人気のある微分積分 >

integral from 0 to 1 of arctan(5x+3)

解

\int_{0}^{1} arctan (5 x + 3) d x

解

\frac{ln ( 10 ) - ln ( 65 ) - 6 arctan ( 3 ) + 16 arctan ( 8 )}{10}

+1

十進法表記

1.37769 \dots

解答ステップ

\int_{0}^{1} arctan (5 x + 3) d x

5 x + 3

の共役で乗じる

\frac{25 x ^{2} - 9}{5 x - 3}

= \int_{0}^{1} arctan (\frac{25 x ^{2} - 9}{5 x - 3}) d x

未定義の (特異) 点を求める:

x = \frac{3}{5}

存在する場合は

b, a < b < c, f (b) =

未定義

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{0}^{\frac{3}{5}} arctan (\frac{25 x ^{2} - 9}{5 x - 3}) d x + \int_{\frac{3}{5}}^{1} arctan (\frac{25 x ^{2} - 9}{5 x - 3}) d x

\int_{0}^{\frac{3}{5}} arctan (\frac{25 x ^{2} - 9}{5 x - 3}) d x = \frac{12 arctan ( 6 ) - ln ( 37 ) - 6 arctan ( 3 ) + ln ( 10 )}{10}

\int_{\frac{3}{5}}^{1} arctan (\frac{25 x ^{2} - 9}{5 x - 3}) d x = \frac{8}{5} arctan (8) - \frac{1}{10} ln (65) - \frac{6}{5} arctan (6) + \frac{1}{10} ln (37)

= \frac{12 arctan ( 6 ) - ln ( 37 ) - 6 arctan ( 3 ) + ln ( 10 )}{10} + \frac{8}{5} arctan (8) - \frac{1}{10} ln (65) - \frac{6}{5} arctan (6) + \frac{1}{10} ln (37)

簡素化

\frac{12 arctan ( 6 ) - ln ( 37 ) - 6 arctan ( 3 ) + ln ( 10 )}{10} + \frac{8}{5} arctan (8) - \frac{1}{10} ln (65) - \frac{6}{5} arctan (6) + \frac{1}{10} ln (37) : \frac{ln ( 10 ) - ln ( 65 ) - 6 arctan ( 3 ) + 16 arctan ( 8 )}{10}

= \frac{ln ( 10 ) - ln ( 65 ) - 6 arctan ( 3 ) + 16 arctan ( 8 )}{10}

グラフ

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