What is the integral of 1/(sqrt(x))+sec^2(x) ?

The integral of 1/(sqrt(x))+sec^2(x) is 2sqrt(x)+tan(x)+C

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integral of 1/(sqrt(x))+sec^2(x)

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

2 x + tan (x) + C

Solution steps

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

\int \frac{1}{x} d x = 2 x

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

= 2 x + tan (x)

Add a constant to the solution

= 2 x + tan (x) + C

(x + a) e^{2 x} + b

\frac{\partial}{\partial y} (6 x y - 2 + 10 x^{2} y)

\frac{d}{d x} (ln (sin^{3} (x^{2} + 7)))

y = tan^{2} (x)

y^{3} + x y^{2} - 61 = x + 3 y^{2}, at x = 3

What is the integral of 1/(sqrt(x))+sec^2(x) ?
The integral of 1/(sqrt(x))+sec^2(x) is 2sqrt(x)+tan(x)+C