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

解

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

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

解答ステップ

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

拡張

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

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

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

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

簡素化

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

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

定数を解答に追加する

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