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

解

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

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

解答ステップ

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

以下の部分分数を得る:

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

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{x}{x ^{2} + 1} d x = \frac{1}{2} ln x^{2} + 1

= ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1

定数を解答に追加する

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