What is the integral of (sec^2(x)+7) ?

The integral of (sec^2(x)+7) is tan(x)+7x+C

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integral of (sec^2(x)+7)

\int (sec^{2} (x) + 7) d x

tan (x) + 7 x + C

Solution steps

\int sec^{2} (x) + 7 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} (x) d x + \int 7 d x

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

\int 7 d x = 7 x

= tan (x) + 7 x

Add a constant to the solution

= tan (x) + 7 x + C

f (x) = - 3 x^{4}, at x = - 1

\int_{- 2}^{3} (x^{2} + 4) - (- x^{2} - 1) d x

x \to \infty lim (8^{- x})

p (x) = 2 x^{4} + 3 x^{3} - 6 x^{2} + 9

\frac{d}{d x} (4 cos (3 x) - 3 sin (4 x))

What is the integral of (sec^2(x)+7) ?
The integral of (sec^2(x)+7) is tan(x)+7x+C