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 e^{x + 7} d x

\frac{\partial}{\partial x} (ln (x^{4}))

\frac{15}{s ^{2} + 2 s + 5}

\int \frac{1}{cos ^{2} ( θ ) 1 + tan ( θ )} d θ

\frac{d x}{d t} = a - b x

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