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人気のある微分積分 >

integral from 1 to 5 of ln(t^2+5t+1)

解

\int_{1}^{5} ln (t^{2} + 5 t + 1) d t

解

\frac{15 ln ( 51 ) + 21 ln ( \frac{5 3}{7} + 1 ) + 21 ln ( \frac{7}{3} - 1 ) - 16 - 7 ln ( 7 ) - 21 ln ( \frac{5 3}{7} - 1 ) - 21 ln ( \frac{7}{3} + 1 )}{2}

+1

十進法表記

12.53418 \dots

解答ステップ

\int_{1}^{5} ln (t^{2} + 5 t + 1) d t

平方完成する

t^{2} + 5 t + 1 : (t + \frac{5}{2})^{2} - \frac{21}{4}

= \int_{1}^{5} ln ((t + \frac{5}{2})^{2} - \frac{21}{4}) d t

u置換積分法を適用する

= \int_{\frac{7}{2}}^{\frac{15}{2}} ln (u^{2} - \frac{21}{4}) d u

部分積分を適用する

= [u ln (u^{2} - \frac{21}{4}) - \int \frac{8 u ^{2}}{4 u ^{2} - 21} d u]_{\frac{7}{2}}^{\frac{15}{2}}

\int \frac{8 u ^{2}}{4 u ^{2} - 21} d u = - \frac{21}{2} ln \frac{2}{21} u + 1 + \frac{21}{2} ln \frac{2}{21} u - 1 + 2 u

= [u ln (u^{2} - \frac{21}{4}) - (- \frac{21}{2} ln \frac{2}{21} u + 1 + \frac{21}{2} ln \frac{2}{21} u - 1 + 2 u)]_{\frac{7}{2}}^{\frac{15}{2}}

簡素化

[u ln (u^{2} - \frac{21}{4}) - (- \frac{21}{2} ln \frac{2}{21} u + 1 + \frac{21}{2} ln \frac{2}{21} u - 1 + 2 u)]_{\frac{7}{2}}^{\frac{15}{2}} : [\frac{1}{2} (2 u ln (u^{2} - \frac{21}{4}) + 21 ln \frac{2}{21} u + 1 - 21 ln \frac{2}{21} u - 1 - 4 u)]_{\frac{7}{2}}^{\frac{15}{2}}

= [\frac{1}{2} (2 u ln (u^{2} - \frac{21}{4}) + 21 ln \frac{2}{21} u + 1 - 21 ln \frac{2}{21} u - 1 - 4 u)]_{\frac{7}{2}}^{\frac{15}{2}}

限度を計算する:

\frac{15 ln ( 51 ) + 21 ln ( \frac{5 3}{7} + 1 ) + 21 ln ( \frac{7}{3} - 1 ) - 16 - 7 ln ( 7 ) - 21 ln ( \frac{5 3}{7} - 1 ) - 21 ln ( \frac{7}{3} + 1 )}{2}

= \frac{15 ln ( 51 ) + 21 ln ( \frac{5 3}{7} + 1 ) + 21 ln ( \frac{7}{3} - 1 ) - 16 - 7 ln ( 7 ) - 21 ln ( \frac{5 3}{7} - 1 ) - 21 ln ( \frac{7}{3} + 1 )}{2}

グラフ

人気の例

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