What is the integral of (6^x-4*15^x)/(3^{2x)} ?

The integral of (6^x-4*15^x)/(3^{2x)} is (2^x)/(ln(2/3)*3^x)-(4*5^x)/(ln(5/3)*3^x)+C

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integral of (6^x-4*15^x)/(3^{2x)}

\int \frac{6 ^{x} - 4 \cdot 1 5 ^{x}}{3 ^{2 x}} d x

\frac{2 ^{x}}{ln ( \frac{2}{3} ) \cdot 3 ^{x}} - \frac{4 \cdot 5 ^{x}}{ln ( \frac{5}{3} ) \cdot 3 ^{x}} + C

Solution steps

\int \frac{6 ^{x} - 4 \cdot 1 5 ^{x}}{3 ^{2 x}} d x

Expand

\frac{6 ^{x} - 4 \cdot 1 5 ^{x}}{3 ^{2 x}} : (\frac{2}{3})^{x} - \frac{4 \cdot 5 ^{x}}{3 ^{x}}

= \int (\frac{2}{3})^{x} - \frac{4 \cdot 5 ^{x}}{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{2}{3})^{x} d x - \int \frac{4 \cdot 5 ^{x}}{3 ^{x}} d x

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

\int \frac{4 \cdot 5 ^{x}}{3 ^{x}} d x = \frac{4 \cdot 5 ^{x}}{ln ( \frac{5}{3} ) \cdot 3 ^{x}}

= \frac{2 ^{x}}{ln ( \frac{2}{3} ) \cdot 3 ^{x}} - \frac{4 \cdot 5 ^{x}}{ln ( \frac{5}{3} ) \cdot 3 ^{x}}

Add a constant to the solution

= \frac{2 ^{x}}{ln ( \frac{2}{3} ) \cdot 3 ^{x}} - \frac{4 \cdot 5 ^{x}}{ln ( \frac{5}{3} ) \cdot 3 ^{x}} + C

ln (ln (6 s))

\frac{d y}{d x} = 3 e^{x - y}

\int \frac{k x}{2} - x^{2} d x

x \to \infty lim (0.5 x^{2})

\frac{\partial}{\partial x} (- 2 cos (2 x))

What is the integral of (6^x-4*15^x)/(3^{2x)} ?
The integral of (6^x-4*15^x)/(3^{2x)} is (2^x)/(ln(2/3)*3^x)-(4*5^x)/(ln(5/3)*3^x)+C