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

解

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

- \frac{4}{x} + \frac{4}{3} x + \frac{x ^{3}}{27} + C

解答ステップ

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

簡素化

(\frac{2}{x} + \frac{x}{3})^{2} : \frac{4}{x ^{2}} + \frac{4}{3} + \frac{x ^{2}}{9}

= \int \frac{4}{x ^{2}} + \frac{4}{3} + \frac{x ^{2}}{9} 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}{3} d x + \int \frac{x ^{2}}{9} d x

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

\int \frac{4}{3} d x = \frac{4}{3} x

\int \frac{x ^{2}}{9} d x = \frac{x ^{3}}{27}

= - \frac{4}{x} + \frac{4}{3} x + \frac{x ^{3}}{27}

定数を解答に追加する

= - \frac{4}{x} + \frac{4}{3} x + \frac{x ^{3}}{27} + C