What is the integral of 3e^x+4sec^2(x) ?

The integral of 3e^x+4sec^2(x) is 3e^x+4tan(x)+C

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integral of 3e^x+4sec^2(x)

\int 3 e^{x} + 4 sec^{2} (x) d x

3 e^{x} + 4 tan (x) + C

Solution steps

\int 3 e^{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 3 e^{x} d x + \int 4 sec^{2} (x) d x

\int 3 e^{x} d x = 3 e^{x}

\int 4 sec^{2} (x) d x = 4 tan (x)

= 3 e^{x} + 4 tan (x)

Add a constant to the solution

= 3 e^{x} + 4 tan (x) + C

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

\int 5 e^{3 x + 1} d x

\int 4 e^{x} + 6 d x

\int \frac{4}{100 - x} d x

\int 5 arccos (x) d x

What is the integral of 3e^x+4sec^2(x) ?
The integral of 3e^x+4sec^2(x) is 3e^x+4tan(x)+C