What is the integral of (2^e+4^{5y}-e)/(2^y) ?

The integral of (2^e+4^{5y}-e)/(2^y) is -(2^{e-y})/(ln(2))+(512^y)/(9ln(2))+e/(ln(2)*2^y)+C

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integral of (2^e+4^{5y}-e)/(2^y)

\int \frac{2 ^{e} + 4 ^{5 y} - e}{2 ^{y}} d y

- \frac{2 ^{e - y}}{ln ( 2 )} + \frac{51 2 ^{y}}{9 ln ( 2 )} + \frac{e}{ln ( 2 ) \cdot 2 ^{y}} + C

Solution steps

\int \frac{2 ^{e} + 4 ^{5 y} - e}{2 ^{y}} d y

\frac{2 ^{e} + 4 ^{5 y} - e}{2 ^{y}} = (\frac{2 ^{e}}{2 ^{y}} + \frac{4 ^{5 y}}{2 ^{y}} - \frac{e}{2 ^{y}})

= \int \frac{2 ^{e}}{2 ^{y}} + \frac{4 ^{5 y}}{2 ^{y}} - \frac{e}{2 ^{y}} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{2 ^{e}}{2 ^{y}} d y + \int \frac{4 ^{5 y}}{2 ^{y}} d y - \int \frac{e}{2 ^{y}} d y

\int \frac{2 ^{e}}{2 ^{y}} d y = - \frac{2 ^{e - y}}{ln ( 2 )}

\int \frac{4 ^{5 y}}{2 ^{y}} d y = \frac{51 2 ^{y}}{9 ln ( 2 )}

\int \frac{e}{2 ^{y}} d y = - \frac{e}{ln ( 2 ) \cdot 2 ^{y}}

= - \frac{2 ^{e - y}}{ln ( 2 )} + \frac{51 2 ^{y}}{9 ln ( 2 )} - (- \frac{e}{ln ( 2 ) \cdot 2 ^{y}})

Simplify

= - \frac{2 ^{e - y}}{ln ( 2 )} + \frac{51 2 ^{y}}{9 ln ( 2 )} + \frac{e}{ln ( 2 ) \cdot 2 ^{y}}

Add a constant to the solution

= - \frac{2 ^{e - y}}{ln ( 2 )} + \frac{51 2 ^{y}}{9 ln ( 2 )} + \frac{e}{ln ( 2 ) \cdot 2 ^{y}} + C

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What is the integral of (2^e+4^{5y}-e)/(2^y) ?
The integral of (2^e+4^{5y}-e)/(2^y) is -(2^{e-y})/(ln(2))+(512^y)/(9ln(2))+e/(ln(2)*2^y)+C