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

The integral of (x^8+sec^2(x)) is (x^9)/9+tan(x)+C

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integral of (x^8+sec^2(x))

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

\frac{x ^{9}}{9} + tan (x) + C

Solution steps

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

\int x^{8} d x = \frac{x ^{9}}{9}

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

= \frac{x ^{9}}{9} + tan (x)

Add a constant to the solution

= \frac{x ^{9}}{9} + tan (x) + C

f (x) = \frac{( x ^{3} + 9 )}{x ^{3}}

x \to 0 lim (tan (3 x))

\int \frac{x - 2}{x} d x

18 t

y = - 2 e^{- x^{2}} x

What is the integral of (x^8+sec^2(x)) ?
The integral of (x^8+sec^2(x)) is (x^9)/9+tan(x)+C