integral of (4x^2+6)/((x^2-2x+2)^2)

解

\int \frac{4 x ^{2} + 6}{( x ^{2} - 2 x + 2 ) ^{2}} d x

\frac{4 x ( x + arctan ( x - 1 ) ( x ^{2} - 2 x + 2 ) - 1 )}{x ^{2} - 2 x + 2} - \frac{1}{2} sin (2 arctan (x - 1)) + 3 arctan (x - 1) - \frac{4 ( ( x - 1 ) ( x ^{2} - 2 x + 2 ) ( \frac{x - 1}{x ^{2} - 2 x + 2} + arctan ( x - 1 ) ) + 1 )}{x ^{2} - 2 x + 2} + C

解答ステップ

\int \frac{4 x ^{2} + 6}{( x ^{2} - 2 x + 2 ) ^{2}} d x

以下の部分分数を得る:

\frac{4 x ^{2} + 6}{( x ^{2} - 2 x + 2 ) ^{2}} : \frac{4}{x ^{2} - 2 x + 2} + \frac{8 x - 2}{( x ^{2} - 2 x + 2 ) ^{2}}

= \int \frac{4}{x ^{2} - 2 x + 2} + \frac{8 x - 2}{( x ^{2} - 2 x + 2 ) ^{2}} d x

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{4}{x ^{2} - 2 x + 2} d x + \int \frac{8 x - 2}{( x ^{2} - 2 x + 2 ) ^{2}} d x

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

\int \frac{8 x - 2}{( x ^{2} - 2 x + 2 ) ^{2}} d x = \frac{4 x ( x + arctan ( x - 1 ) ( x ^{2} - 2 x + 2 ) - 1 )}{x ^{2} - 2 x + 2} - \frac{4 ( ( x - 1 ) ( \frac{x - 1}{x ^{2} - 2 x + 2} + arctan ( x - 1 ) ) ( x ^{2} - 2 x + 2 ) + 1 )}{x ^{2} - 2 x + 2} - arctan (x - 1) - \frac{1}{2} sin (2 arctan (x - 1))

= 4 arctan (x - 1) + \frac{4 x ( x + arctan ( x - 1 ) ( x ^{2} - 2 x + 2 ) - 1 )}{x ^{2} - 2 x + 2} - \frac{4 ( ( x - 1 ) ( \frac{x - 1}{x ^{2} - 2 x + 2} + arctan ( x - 1 ) ) ( x ^{2} - 2 x + 2 ) + 1 )}{x ^{2} - 2 x + 2} - arctan (x - 1) - \frac{1}{2} sin (2 arctan (x - 1))

簡素化

= \frac{4 x ( x + arctan ( x - 1 ) ( x ^{2} - 2 x + 2 ) - 1 )}{x ^{2} - 2 x + 2} - \frac{1}{2} sin (2 arctan (x - 1)) + 3 arctan (x - 1) - \frac{4 ( ( x - 1 ) ( x ^{2} - 2 x + 2 ) ( \frac{x - 1}{x ^{2} - 2 x + 2} + arctan ( x - 1 ) ) + 1 )}{x ^{2} - 2 x + 2}

定数を解答に追加する

= \frac{4 x ( x + arctan ( x - 1 ) ( x ^{2} - 2 x + 2 ) - 1 )}{x ^{2} - 2 x + 2} - \frac{1}{2} sin (2 arctan (x - 1)) + 3 arctan (x - 1) - \frac{4 ( ( x - 1 ) ( x ^{2} - 2 x + 2 ) ( \frac{x - 1}{x ^{2} - 2 x + 2} + arctan ( x - 1 ) ) + 1 )}{x ^{2} - 2 x + 2} + C