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

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

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integral of sec(x)tan(x)+4x

\int sec (x) tan (x) + 4 x d x

sec (x) + 2 x^{2} + C

Solution steps

\int sec (x) tan (x) + 4 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 sec (x) tan (x) d x + \int 4 x d x

\int sec (x) tan (x) d x = sec (x)

\int 4 x d x = 2 x^{2}

= sec (x) + 2 x^{2}

Add a constant to the solution

= sec (x) + 2 x^{2} + C

n = 0 \sum \infty (\frac{1}{4})^{n}

\frac{\partial}{\partial x} (e^{x y z^{8}})

\frac{\partial}{\partial x} (6 y^{2})

f (x) = (3 x^{2} - 5 x) e^{x}

f (t) = (t^{62} + 3 t + 3) (5 t^{2} + 4)

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