integral from 0 to 1 of 1/(3x^2-6x+6)

解

\int_{0}^{1} \frac{1}{3 x ^{2} - 6 x + 6} d x

\frac{π}{12}

十進法表記

0.26179 \dots

解答ステップ

\int_{0}^{1} \frac{1}{3 x ^{2} - 6 x + 6} d x

部分積分を適用する

= [\frac{x}{3 x ^{2} - 6 x + 6} - \int - \frac{2 x ( x - 1 )}{3 ( x ^{2} - 2 x + 2 ) ^{2}} d x]_{0}^{1}

\int - \frac{2 x ( x - 1 )}{3 ( x ^{2} - 2 x + 2 ) ^{2}} d x = - \frac{x ( x + arctan ( x - 1 ) ( x ^{2} - 2 x + 2 ) - 1 )}{3 ( x ^{2} - 2 x + 2 )} + \frac{1}{3} sin (2 arctan (x - 1)) + \frac{( x - 1 ) ( x ^{2} - 2 x + 2 ) ( \frac{x - 1}{x ^{2} - 2 x + 2} + arctan ( x - 1 ) ) + 1}{3 ( x ^{2} - 2 x + 2 )}

= [\frac{x}{3 x ^{2} - 6 x + 6} - (- \frac{x ( x + arctan ( x - 1 ) ( x ^{2} - 2 x + 2 ) - 1 )}{3 ( x ^{2} - 2 x + 2 )} + \frac{1}{3} sin (2 arctan (x - 1)) + \frac{( x - 1 ) ( x ^{2} - 2 x + 2 ) ( \frac{x - 1}{x ^{2} - 2 x + 2} + arctan ( x - 1 ) ) + 1}{3 ( x ^{2} - 2 x + 2 )})]_{0}^{1}

簡素化

[\frac{x}{3 x ^{2} - 6 x + 6} - (- \frac{x ( x + arctan ( x - 1 ) ( x ^{2} - 2 x + 2 ) - 1 )}{3 ( x ^{2} - 2 x + 2 )} + \frac{1}{3} sin (2 arctan (x - 1)) + \frac{( x - 1 ) ( x ^{2} - 2 x + 2 ) ( \frac{x - 1}{x ^{2} - 2 x + 2} + arctan ( x - 1 ) ) + 1}{3 ( x ^{2} - 2 x + 2 )})]_{0}^{1} : [\frac{x ^{4} arctan ( x - 1 ) - x ^{4} sin ( 2 arctan ( x - 1 ) ) + 2 x ^{3} + 4 x ^{3} sin ( 2 arctan ( x - 1 ) ) - 4 x ^{3} arctan ( x - 1 ) + 8 x ^{2} arctan ( x - 1 ) - 6 x ^{2} - 8 x ^{2} sin ( 2 arctan ( x - 1 ) ) + 8 x sin ( 2 arctan ( x - 1 ) ) + 8 x - 8 x arctan ( x - 1 ) + 4 arctan ( x - 1 ) - 4 sin ( 2 arctan ( x - 1 ) ) - 4}{3 ( x ^{2} - 2 x + 2 ) ^{2}}]_{0}^{1}

= [\frac{x ^{4} arctan ( x - 1 ) - x ^{4} sin ( 2 arctan ( x - 1 ) ) + 2 x ^{3} + 4 x ^{3} sin ( 2 arctan ( x - 1 ) ) - 4 x ^{3} arctan ( x - 1 ) + 8 x ^{2} arctan ( x - 1 ) - 6 x ^{2} - 8 x ^{2} sin ( 2 arctan ( x - 1 ) ) + 8 x sin ( 2 arctan ( x - 1 ) ) + 8 x - 8 x arctan ( x - 1 ) + 4 arctan ( x - 1 ) - 4 sin ( 2 arctan ( x - 1 ) ) - 4}{3 ( x ^{2} - 2 x + 2 ) ^{2}}]_{0}^{1}

限度を計算する:

\frac{π}{12}

= \frac{π}{12}