integral of 1/((2x-1)^2(x^2-2x+2))

解

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

\frac{4}{25} ln ∣ 2 x - 1 ∣ - \frac{2}{5 ( 2 x - 1 )} + \frac{1}{25} (- 2 ln x^{2} - 2 x + 2 - 3 arctan (x - 1)) + C

解答ステップ

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

以下の部分分数を得る:

\frac{1}{( 2 x - 1 ) ^{2} ( x ^{2} - 2 x + 2 )} : \frac{8}{25 ( 2 x - 1 )} + \frac{4}{5 ( 2 x - 1 ) ^{2}} + \frac{- 4 x + 1}{25 ( x ^{2} - 2 x + 2 )}

= \int \frac{8}{25 ( 2 x - 1 )} + \frac{4}{5 ( 2 x - 1 ) ^{2}} + \frac{- 4 x + 1}{25 ( x ^{2} - 2 x + 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{8}{25 ( 2 x - 1 )} d x + \int \frac{4}{5 ( 2 x - 1 ) ^{2}} d x + \int \frac{- 4 x + 1}{25 ( x ^{2} - 2 x + 2 )} d x

\int \frac{8}{25 ( 2 x - 1 )} d x = \frac{4}{25} ln ∣ 2 x - 1 ∣

\int \frac{4}{5 ( 2 x - 1 ) ^{2}} d x = - \frac{2}{5 ( 2 x - 1 )}

\int \frac{- 4 x + 1}{25 ( x ^{2} - 2 x + 2 )} d x = \frac{1}{25} (- 2 ln x^{2} - 2 x + 2 - 3 arctan (x - 1))

= \frac{4}{25} ln ∣ 2 x - 1 ∣ - \frac{2}{5 ( 2 x - 1 )} + \frac{1}{25} (- 2 ln x^{2} - 2 x + 2 - 3 arctan (x - 1))

定数を解答に追加する

= \frac{4}{25} ln ∣ 2 x - 1 ∣ - \frac{2}{5 ( 2 x - 1 )} + \frac{1}{25} (- 2 ln x^{2} - 2 x + 2 - 3 arctan (x - 1)) + C