What is the integral of sec^2(x)-1 ?

The integral of sec^2(x)-1 is tan(x)-x+C

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integral of sec^2(x)-1

\int sec^{2} (x) - 1 d x

tan (x) - x + C

Solution steps

\int sec^{2} (x) - 1 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 1 d x

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

\int 1 d x = x

= tan (x) - x

Add a constant to the solution

= tan (x) - x + C

y = \frac{x}{x + 1}

x \to - 9 + lim (\frac{x + 8}{x + 9})

\int \frac{1}{4 x ^{2} + 4 ^{2}} d x

f (x) = sin (5) e^{- 5 x}

x \to 0 lim (\frac{1}{x - 1})