What is the integral of 4^x+ln(5^x) ?

The integral of 4^x+ln(5^x) is (2^{2x-1})/(ln(2))+1/2 ln(5)x^2+C

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integral of 4^x+ln(5^x)

\int 4^{x} + ln (5^{x}) d x

\frac{2 ^{2 x - 1}}{ln ( 2 )} + \frac{1}{2} ln (5) x^{2} + C

Solution steps

\int 4^{x} + ln (5^{x}) d x

Apply log rule

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x),

assuming

x \geq 0

= \int 4^{x} + ln (5) 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 ln (5) x d x

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

\int ln (5) x d x = \frac{1}{2} ln (5) x^{2}

= \frac{2 ^{2 x - 1}}{ln ( 2 )} + \frac{1}{2} ln (5) x^{2}

Add a constant to the solution

= \frac{2 ^{2 x - 1}}{ln ( 2 )} + \frac{1}{2} ln (5) x^{2} + C

\frac{3}{s ^{2} + 7}

\int 8 x d x

y = 16 x, (16, 64)

\frac{d}{d x} (arctan (\frac{3}{x + 2}))

\int \frac{sin ( x )}{2 + cos ( x )} d x

What is the integral of 4^x+ln(5^x) ?
The integral of 4^x+ln(5^x) is (2^{2x-1})/(ln(2))+1/2 ln(5)x^2+C