What is the integral of (3sec^2(x)+1/(x^2)+2^x-4) ?

The integral of (3sec^2(x)+1/(x^2)+2^x-4) is 3tan(x)-1/x+(2^x)/(ln(2))-4x+C

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integral of (3sec^2(x)+1/(x^2)+2^x-4)

\int (3 sec^{2} (x) + \frac{1}{x ^{2}} + 2^{x} - 4) d x

3 tan (x) - \frac{1}{x} + \frac{2 ^{x}}{ln ( 2 )} - 4 x + C

Solution steps

\int 3 sec^{2} (x) + \frac{1}{x ^{2}} + 2^{x} - 4 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 sec^{2} (x) d x + \int \frac{1}{x ^{2}} d x + \int 2^{x} d x - \int 4 d x

\int 3 sec^{2} (x) d x = 3 tan (x)

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

\int 2^{x} d x = \frac{2 ^{x}}{ln ( 2 )}

\int 4 d x = 4 x

= 3 tan (x) - \frac{1}{x} + \frac{2 ^{x}}{ln ( 2 )} - 4 x

Add a constant to the solution

= 3 tan (x) - \frac{1}{x} + \frac{2 ^{x}}{ln ( 2 )} - 4 x + C

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What is the integral of (3sec^2(x)+1/(x^2)+2^x-4) ?
The integral of (3sec^2(x)+1/(x^2)+2^x-4) is 3tan(x)-1/x+(2^x)/(ln(2))-4x+C