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

解

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

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

解答ステップ

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

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int \frac{1}{x ( x ^{2} + 1 )} d x

以下の部分分数を得る:

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

= 3 \cdot \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

= 3 (\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

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

定数を解答に追加する

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