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

解

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

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

解答ステップ

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

拡張

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

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

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

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

定数を解答に追加する

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