integral of (4x^2-3x+3)/((x-1)(x^2+1))

解

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

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

解答ステップ

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

拡張

\frac{4 x ^{2} - 3 x + 3}{( x - 1 ) ( x ^{2} + 1 )} : \frac{4 x ^{2}}{( x - 1 ) ( x ^{2} + 1 )} - \frac{3 x}{( x - 1 ) ( x ^{2} + 1 )} + \frac{3}{( x - 1 ) ( 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 ) ( x ^{2} + 1 )} d x - \int \frac{3 x}{( x - 1 ) ( x ^{2} + 1 )} d x + \int \frac{3}{( x - 1 ) ( x ^{2} + 1 )} d x

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

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

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

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

簡素化

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

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

定数を解答に追加する

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