What is the integral of sec^2(x)+5 ?

The integral of sec^2(x)+5 is tan(x)+5x+C

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integral of sec^2(x)+5

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

tan (x) + 5 x + C

Solution steps

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

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

\int 5 d x = 5 x

= tan (x) + 5 x

Add a constant to the solution

= tan (x) + 5 x + C

f (x) = 2 x^{3} + 5 x^{2} - 2, at x = - 2

\frac{( 3 t - 2 ) ^{3}}{t + 5}

\frac{( e ^{- x} )}{x}

\frac{d}{d x} (cos (arccos (a x^{2} + b x + c)))

\frac{\partial}{\partial x} ((e^{x} - 4 x) sin (y))