What is the integral of sec^2(5x)+e^{x/2} ?

The integral of sec^2(5x)+e^{x/2} is 1/5 tan(5x)+2e^{x/2}+C

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integral of sec^2(5x)+e^{x/2}

\int sec^{2} (5 x) + e^{\frac{x}{2}} d x

\frac{1}{5} tan (5 x) + 2 e^{\frac{x}{2}} + C

Solution steps

\int sec^{2} (5 x) + e^{\frac{x}{2}} 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 sec^{2} (5 x) d x + \int e^{\frac{x}{2}} d x

\int sec^{2} (5 x) d x = \frac{1}{5} tan (5 x)

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

= \frac{1}{5} tan (5 x) + 2 e^{\frac{x}{2}}

Add a constant to the solution

= \frac{1}{5} tan (5 x) + 2 e^{\frac{x}{2}} + C

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

y^{^{'}} (t) = (y + 1) (y - 3), y (0) = 1

t^{3} \frac{d y}{d t} + 3 t^{2} y = 7 cos (t), y (π) = 0

x \to \infty lim (- \frac{x}{s})

18 (2 e^{x^{2}} x^{2} + e^{x^{2}})

What is the integral of sec^2(5x)+e^{x/2} ?
The integral of sec^2(5x)+e^{x/2} is 1/5 tan(5x)+2e^{x/2}+C