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

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

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integral of 1+sec^2(x)

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

x + tan (x) + C

Solution steps

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

\int 1 d x = x

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

= x + tan (x)

Add a constant to the solution

= x + tan (x) + C

sec^{2} (x) - 5

n = 0 \sum \infty 0. 2^{n}

\frac{d}{d x} (- \frac{x ^{l n (x)}}{6})

f (x) = x^{2} - 4 x, g (x) = 0