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

解答

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

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

求解步骤

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

使用积分加法定则:

\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}

解答补常数

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

\frac{d}{d x} (e^{x y} cos (x + y))

y^{^{'}} - \frac{3}{x ^{2}} y = \frac{1}{x ^{2}}

y^{^{'}} = \frac{t}{t ^{2} y + y}

n = 1 \sum \infty n e^{- \frac{n}{8}}

y^{^{''}} + 20 y^{^{'}} = 0, y (0) = - 5, y^{^{'}} (0) = 20