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

The integral of (4^x-1/(3x)) is (2^{2x-1})/(ln(2))-1/3 ln|x|+C

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integral of (4^x-1/(3x))

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

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

Solution steps

\int 4^{x} - \frac{1}{3 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 4^{x} d x - \int \frac{1}{3 x} d x

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

\int \frac{1}{3 x} d x = \frac{1}{3} ln ∣ x ∣

= \frac{2 ^{2 x - 1}}{ln ( 2 )} - \frac{1}{3} ln ∣ x ∣

Add a constant to the solution

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

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