What is the integral of ((3^x-2^x)^2)/(6^x) ?

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

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integral of ((3^x-2^x)^2)/(6^x)

\int \frac{( 3 ^{x} - 2 ^{x} ) ^{2}}{6 ^{x}} d x

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

Solution steps

\int \frac{( 3 ^{x} - 2 ^{x} ) ^{2}}{6 ^{x}} d x

Expand

\frac{( 3 ^{x} - 2 ^{x} ) ^{2}}{6 ^{x}} : (\frac{3}{2})^{x} - 2 + (\frac{2}{3})^{x}

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

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

\int 2 d x = 2 x

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

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

Add a constant to the solution

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

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