What is the integral of 3/x+10sec^2(x)+40e^x ?

The integral of 3/x+10sec^2(x)+40e^x is 3ln|x|+10tan(x)+40e^x+C

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

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

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

Solution steps

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

Apply the Sum Rule:

\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}

Add a constant to the solution

= 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

What is the integral of 3/x+10sec^2(x)+40e^x ?
The integral of 3/x+10sec^2(x)+40e^x is 3ln|x|+10tan(x)+40e^x+C