What is the integral of (x^4+sec^2(x)) ?

The integral of (x^4+sec^2(x)) is (x^5)/5+tan(x)+C

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integral of (x^4+sec^2(x))

\int (x^{4} + sec^{2} (x)) d x

\frac{x ^{5}}{5} + tan (x) + C

Solution steps

\int x^{4} + 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^{4} d x + \int sec^{2} (x) d x

\int x^{4} d x = \frac{x ^{5}}{5}

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

= \frac{x ^{5}}{5} + tan (x)

Add a constant to the solution

= \frac{x ^{5}}{5} + tan (x) + C

\frac{d}{d x} (x e^{y x})

\frac{d}{d x} (e^{- x} cos (3 x))

\int_{1}^{3} t ln (t) d t

(\frac{1}{5}) (x y^{2} + 4 y)^{2}

What is the integral of (x^4+sec^2(x)) ?
The integral of (x^4+sec^2(x)) is (x^5)/5+tan(x)+C