integral of (4x-5)/((x-1)(x^2+1))

解

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

- \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{4} ln x^{2} + 1 + \frac{9}{2} arctan (x) + C

解答ステップ

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

拡張

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

\int \frac{4 x}{( 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{5}{( x - 1 ) ( x ^{2} + 1 )} d x = 5 (\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))) - 5 (\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))) - 5 (\frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} (- \frac{1}{2} ln x^{2} + 1 - arctan (x))) : - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{4} ln x^{2} + 1 + \frac{9}{2} arctan (x)

= - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{4} ln x^{2} + 1 + \frac{9}{2} arctan (x)

定数を解答に追加する

= - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{4} ln x^{2} + 1 + \frac{9}{2} arctan (x) + C