integral of (4x^2-8x)/((x-1)^2(x^2+1))

解

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

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

解答ステップ

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

拡張

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

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}}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x - \int \frac{8 x}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x

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

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

= 2 ln ∣ x - 1 ∣ - \frac{2}{x - 1} - ln x^{2} + 1 - 8 (- \frac{1}{2 ( x - 1 )} - \frac{1}{2} arctan (x))

簡素化

2 ln ∣ x - 1 ∣ - \frac{2}{x - 1} - ln x^{2} + 1 - 8 (- \frac{1}{2 ( x - 1 )} - \frac{1}{2} arctan (x)) : 2 ln ∣ x - 1 ∣ + \frac{2}{x - 1} - ln x^{2} + 1 + 4 arctan (x)

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

定数を解答に追加する

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