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

解

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

- \frac{1}{x} + 4 x + \frac{x ^{2}}{2} + C

解答ステップ

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

拡張

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

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

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

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

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

= - \frac{1}{x} + 4 x^{\frac{1}{2}} + \frac{x ^{2}}{2}

累乗根の規則を適用する:

a^{\frac{1}{n}} = n a

= - \frac{1}{x} + 4 x + \frac{x ^{2}}{2}

定数を解答に追加する

= - \frac{1}{x} + 4 x + \frac{x ^{2}}{2} + C