What is the integral of (-sec^2(x)+e^{2x+1}) ?

The integral of (-sec^2(x)+e^{2x+1}) is -tan(x)+1/2 e^{2x+1}+C

integral of (-sec^2(x)+e^{2x+1})

\int (- sec^{2} (x) + e^{2 x + 1}) d x

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

Solution steps

\int - sec^{2} (x) + e^{2 x + 1} 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^{2} (x) d x + \int e^{2 x + 1} d x

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

\int e^{2 x + 1} d x = \frac{1}{2} e^{2 x + 1}

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

Add a constant to the solution

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

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What is the integral of (-sec^2(x)+e^{2x+1}) ?
The integral of (-sec^2(x)+e^{2x+1}) is -tan(x)+1/2 e^{2x+1}+C