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

The integral of (2^x-3^x)^2 is (2^{2x-1})/(ln(2))-(3^x*2^{x+1})/(ln(6))+(9^x)/(2ln(3))+C

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

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

\frac{2 ^{2 x - 1}}{ln ( 2 )} - \frac{3 ^{x} \cdot 2 ^{x + 1}}{ln ( 6 )} + \frac{9 ^{x}}{2 ln ( 3 )} + C

Solution steps

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

Expand

(2^{x} - 3^{x})^{2} : 4^{x} - 3^{x} \cdot 2^{1 + x} + 9^{x}

= \int 4^{x} - 3^{x} \cdot 2^{1 + x} + 9^{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 4^{x} d x - \int 3^{x} \cdot 2^{1 + x} d x + \int 9^{x} d x

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

\int 3^{x} \cdot 2^{1 + x} d x = \frac{3 ^{x} \cdot 2 ^{x + 1}}{ln ( 6 )}

\int 9^{x} d x = \frac{9 ^{x}}{2 ln ( 3 )}

= \frac{2 ^{2 x - 1}}{ln ( 2 )} - \frac{3 ^{x} \cdot 2 ^{x + 1}}{ln ( 6 )} + \frac{9 ^{x}}{2 ln ( 3 )}

Add a constant to the solution

= \frac{2 ^{2 x - 1}}{ln ( 2 )} - \frac{3 ^{x} \cdot 2 ^{x + 1}}{ln ( 6 )} + \frac{9 ^{x}}{2 ln ( 3 )} + C

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What is the integral of (2^x-3^x)^2 ?
The integral of (2^x-3^x)^2 is (2^{2x-1})/(ln(2))-(3^x*2^{x+1})/(ln(6))+(9^x)/(2ln(3))+C