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

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

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integral of 9+9tan^2(x)

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

9 tan (x) + C

Solution steps

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

9 + 9 tan^{2} (x) = 9 (1 + tan^{2} (x))

= \int 9 (1 + tan^{2} (x)) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \int 1 + tan^{2} (x) d x

Rewrite using trig identities

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

Use the common integral:

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

= 9 tan (x)

Add a constant to the solution

= 9 tan (x) + C

\frac{d}{d x} (x (7 x + 9)^{\frac{1}{3}})

\int \frac{3 x ^{2}}{4} d x

\frac{\partial}{\partial x} (e^{(x - 1)^{2}})

x^{4} + x - 2

25 y^{^{''}} + 2 y = 0