integral of ((e^{1/x}+2))/(x^2)

解

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

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

解答ステップ

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

u置換積分法を適用する

= \int - e^{u} - 2 d u

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 e^{u} d u - \int 2 d u

\int e^{u} d u = e^{u}

\int 2 d u = 2 u

= - e^{u} - 2 u

代用を戻す

u = \frac{1}{x}

= - e^{\frac{1}{x}} - 2 \cdot \frac{1}{x}

2 \cdot \frac{1}{x} = \frac{2}{x}

= - e^{\frac{1}{x}} - \frac{2}{x}

定数を解答に追加する

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