What is the integral of (2^x-3^x)/(5^x) ?

The integral of (2^x-3^x)/(5^x) is (2^x)/(ln(2/5)*5^x)-(3^x)/(ln(3/5)*5^x)+C

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integral of (2^x-3^x)/(5^x)

\int \frac{2 ^{x} - 3 ^{x}}{5 ^{x}} d x

\frac{2 ^{x}}{ln ( \frac{2}{5} ) \cdot 5 ^{x}} - \frac{3 ^{x}}{ln ( \frac{3}{5} ) \cdot 5 ^{x}} + C

Solution steps

\int \frac{2 ^{x} - 3 ^{x}}{5 ^{x}} d x

Expand

\frac{2 ^{x} - 3 ^{x}}{5 ^{x}} : (\frac{2}{5})^{x} - (\frac{3}{5})^{x}

= \int (\frac{2}{5})^{x} - (\frac{3}{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{2}{5})^{x} d x - \int (\frac{3}{5})^{x} d x

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

\int (\frac{3}{5})^{x} d x = \frac{3 ^{x}}{ln ( \frac{3}{5} ) \cdot 5 ^{x}}

= \frac{2 ^{x}}{ln ( \frac{2}{5} ) \cdot 5 ^{x}} - \frac{3 ^{x}}{ln ( \frac{3}{5} ) \cdot 5 ^{x}}

Add a constant to the solution

= \frac{2 ^{x}}{ln ( \frac{2}{5} ) \cdot 5 ^{x}} - \frac{3 ^{x}}{ln ( \frac{3}{5} ) \cdot 5 ^{x}} + C

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What is the integral of (2^x-3^x)/(5^x) ?
The integral of (2^x-3^x)/(5^x) is (2^x)/(ln(2/5)*5^x)-(3^x)/(ln(3/5)*5^x)+C