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

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

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integral of (x^2+4^x)

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

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

Solution steps

\int x^{2} + 4^{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 x^{2} d x + \int 4^{x} d x

\int x^{2} d x = \frac{x ^{3}}{3}

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

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

Add a constant to the solution

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

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