What is the integral of 10x^4-sec^2(x) ?

The integral of 10x^4-sec^2(x) is 2x^5-tan(x)+C

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integral of 10x^4-sec^2(x)

\int 10 x^{4} - sec^{2} (x) d x

2 x^{5} - tan (x) + C

Solution steps

\int 10 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 10 x^{4} d x - \int sec^{2} (x) d x

\int 10 x^{4} d x = 2 x^{5}

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

= 2 x^{5} - tan (x)

Add a constant to the solution

= 2 x^{5} - tan (x) + C

\frac{d}{d x} (x^{3} (3 x^{6} - 5 x^{2} + 4))

y = x, y = x, x = 4

\int \frac{1}{3} (1 - x^{2})^{3} d x

cos (x) \frac{d y}{d x} + (sin (x)) y = 1

\frac{\partial}{\partial y} (e^{x^{2} + y^{2}} sin (x y))

What is the integral of 10x^4-sec^2(x) ?
The integral of 10x^4-sec^2(x) is 2x^5-tan(x)+C