What is the integral of 5^x-sec^2(x) ?

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

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integral of 5^x-sec^2(x)

\int 5^{x} - sec^{2} (x) d x

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

Solution steps

\int 5^{x} - 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 5^{x} d x - \int sec^{2} (x) d x

\int 5^{x} d x = \frac{5 ^{x}}{ln ( 5 )}

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

= \frac{5 ^{x}}{ln ( 5 )} - tan (x)

Add a constant to the solution

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

- x^{2} - 1, x^{2} + 4, [- 2, 3]

y^{^{'}} = x (1 + y)

\frac{\partial}{\partial x} (t x)

\int - \frac{1}{2} x d x

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