What is the integral of 1/2 sec^2(x)+4x ?

The integral of 1/2 sec^2(x)+4x is 1/2 tan(x)+2x^2+C

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integral of 1/2 sec^2(x)+4x

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

\frac{1}{2} tan (x) + 2 x^{2} + C

Solution steps

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

\int \frac{1}{2} sec^{2} (x) d x = \frac{1}{2} tan (x)

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

= \frac{1}{2} tan (x) + 2 x^{2}

Add a constant to the solution

= \frac{1}{2} tan (x) + 2 x^{2} + C

\frac{d ^{3}}{d x ^{3}} (2 x + 8)

\int_{- 5}^{0} (25 - x^{2}) d x

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

\frac{\partial}{\partial z} (x y + z^{3})

y = 2 sin (x), y = 2 cos (x)

What is the integral of 1/2 sec^2(x)+4x ?
The integral of 1/2 sec^2(x)+4x is 1/2 tan(x)+2x^2+C