integral of (2/x+1)^2

解

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

- \frac{4}{x} + 4 ln ∣ x ∣ + x + C

解答ステップ

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

拡張

(\frac{2}{x} + 1)^{2} : \frac{4}{x ^{2}} + \frac{4}{x} + 1

= \int \frac{4}{x ^{2}} + \frac{4}{x} + 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{4}{x ^{2}} d x + \int \frac{4}{x} d x + \int 1 d x

\int \frac{4}{x ^{2}} d x = - \frac{4}{x}

\int \frac{4}{x} d x = 4 ln ∣ x ∣

\int 1 d x = x

= - \frac{4}{x} + 4 ln ∣ x ∣ + x

定数を解答に追加する

= - \frac{4}{x} + 4 ln ∣ x ∣ + x + C