integral of 3/x+10sec^2(x)+40e^x

解

\int \frac{3}{x} + 10 sec^{2} (x) + 40 e^{x} d x

3 ln ∣ x ∣ + 10 tan (x) + 40 e^{x} + C

解答ステップ

\int \frac{3}{x} + 10 sec^{2} (x) + 40 e^{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{3}{x} d x + \int 10 sec^{2} (x) d x + \int 40 e^{x} d x

\int \frac{3}{x} d x = 3 ln ∣ x ∣

\int 10 sec^{2} (x) d x = 10 tan (x)

\int 40 e^{x} d x = 40 e^{x}

= 3 ln ∣ x ∣ + 10 tan (x) + 40 e^{x}

定数を解答に追加する

= 3 ln ∣ x ∣ + 10 tan (x) + 40 e^{x} + C