What is the integral of sec^2(x)+9 ?

The integral of sec^2(x)+9 is tan(x)+9x+C

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integral of sec^2(x)+9

\int sec^{2} (x) + 9 d x

tan (x) + 9 x + C

Solution steps

\int sec^{2} (x) + 9 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 9 d x

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

\int 9 d x = 9 x

= tan (x) + 9 x

Add a constant to the solution

= tan (x) + 9 x + C

\int - \frac{5 ^{- x}}{ln ( 5 )} d x

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\int (x^{2} - 5 x + 3) d x

y^{^{''}} + 2 y^{^{'}} - 8 y = 2 e^{- 3 x} - e^{- x}

n = 0 \sum \infty (x - 2)^{n}