What is the integral of (3^x-2^{2x})/(e^x) ?

The integral of (3^x-2^{2x})/(e^x) is (3^x)/((ln(3)-1)e^x)-(4^x)/((2ln(2)-1)e^x)+C

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integral of (3^x-2^{2x})/(e^x)

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

\frac{3 ^{x}}{( ln ( 3 ) - 1 ) e ^{x}} - \frac{4 ^{x}}{( 2 ln ( 2 ) - 1 ) e ^{x}} + C

Solution steps

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

Expand

\frac{3 ^{x} - 2 ^{2 x}}{e ^{x}} : (\frac{3}{e})^{x} - (\frac{4}{e})^{x}

= \int (\frac{3}{e})^{x} - (\frac{4}{e})^{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{3}{e})^{x} d x - \int (\frac{4}{e})^{x} d x

\int (\frac{3}{e})^{x} d x = \frac{3 ^{x}}{( ln ( 3 ) - 1 ) e ^{x}}

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

= \frac{3 ^{x}}{( ln ( 3 ) - 1 ) e ^{x}} - \frac{4 ^{x}}{( 2 ln ( 2 ) - 1 ) e ^{x}}

Add a constant to the solution

= \frac{3 ^{x}}{( ln ( 3 ) - 1 ) e ^{x}} - \frac{4 ^{x}}{( 2 ln ( 2 ) - 1 ) e ^{x}} + C

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What is the integral of (3^x-2^{2x})/(e^x) ?
The integral of (3^x-2^{2x})/(e^x) is (3^x)/((ln(3)-1)e^x)-(4^x)/((2ln(2)-1)e^x)+C