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

The integral of 8+tan^2(x) is 7x+tan(x)+C

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integral of 8+tan^2(x)

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

7 x + tan (x) + C

Solution steps

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

\int 8 d x = 8 x

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

= 8 x - x + tan (x)

Simplify

= 7 x + tan (x)

Add a constant to the solution

= 7 x + tan (x) + C

\frac{d}{d x} (sin^{2} (3 + x))

\frac{d y}{d x} = (\frac{( 4 y + 3 )}{8 x + 5})^{2}

x \to 6 lim (4 x + ∣ x - 6 ∣)

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

x \to 2 lim (\frac{2 x + 1}{x + 2})