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

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

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

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

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

Solution steps

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

Apply log rule

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

assuming

x \geq 0

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

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^{u} d u + \int ln (5) u d u

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

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

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

Add a constant to the solution

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

\frac{d}{d x} (y^{3 x})

\int (x^{2} - 3)^{3} (2 x) d x

f (x) = 7 - 5 x^{2}

\int 3 x + 1 d x

f (x) = x^{2} - 2, (5, 23)

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