What is the integral of (2^{x+1}-5^{x-1})/(10^x) ?

The integral of (2^{x+1}-5^{x-1})/(10^x) is -2/(ln(5)*5^x)+1/(5ln(2)*2^x)+C

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integral of (2^{x+1}-5^{x-1})/(10^x)

\int \frac{2 ^{x + 1} - 5 ^{x - 1}}{1 0 ^{x}} d x

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

Solution steps

\int \frac{2 ^{x + 1} - 5 ^{x - 1}}{1 0 ^{x}} d x

\frac{2 ^{x + 1} - 5 ^{x - 1}}{1 0 ^{x}} = (\frac{2 ^{x + 1}}{1 0 ^{x}} - \frac{5 ^{x - 1}}{1 0 ^{x}})

= \int \frac{2 ^{x + 1}}{1 0 ^{x}} - \frac{5 ^{x - 1}}{1 0 ^{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 \frac{2 ^{x + 1}}{1 0 ^{x}} d x - \int \frac{5 ^{x - 1}}{1 0 ^{x}} d x

\int \frac{2 ^{x + 1}}{1 0 ^{x}} d x = - \frac{2}{ln ( 5 ) \cdot 5 ^{x}}

\int \frac{5 ^{x - 1}}{1 0 ^{x}} d x = - \frac{1}{5 ln ( 2 ) \cdot 2 ^{x}}

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

Simplify

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

Add a constant to the solution

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

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x \to 2 lim (\frac{x ^{3} - 6 x ^{2} + 12 x - 8}{x ^{4} - 4 x ^{3} + 16 x - 16})

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\frac{d}{d x} (a e^{x} + b e^{- 9 x})

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What is the integral of (2^{x+1}-5^{x-1})/(10^x) ?
The integral of (2^{x+1}-5^{x-1})/(10^x) is -2/(ln(5)*5^x)+1/(5ln(2)*2^x)+C