What is the integral of sec^2(x)+8 ?

The integral of sec^2(x)+8 is tan(x)+8x+C

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integral of sec^2(x)+8

\int sec^{2} (x) + 8 d x

tan (x) + 8 x + C

Solution steps

\int sec^{2} (x) + 8 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 8 d x

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

\int 8 d x = 8 x

= tan (x) + 8 x

Add a constant to the solution

= tan (x) + 8 x + C

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