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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

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