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

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

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integral of (32^x-23^x)/(2x)

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{3 \cdot 2 ^{x} - 2 \cdot 3 ^{x}}{x} d x

Expand

\frac{3 \cdot 2 ^{x} - 2 \cdot 3 ^{x}}{x} : \frac{3 \cdot 2 ^{x}}{x} - \frac{2 \cdot 3 ^{x}}{x}

= \frac{1}{2} \cdot \int \frac{3 \cdot 2 ^{x}}{x} - \frac{2 \cdot 3 ^{x}}{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

= \frac{1}{2} (\int \frac{3 \cdot 2 ^{x}}{x} d x - \int \frac{2 \cdot 3 ^{x}}{x} d x)

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

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

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

Simplify

\frac{1}{2} (3 Ei (ln (2^{x})) - 2 Ei (ln (3^{x}))) : \frac{3}{2} Ei (ln (2^{x})) - Ei (ln (3^{x}))

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

Add a constant to the solution

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

\frac{1}{x}

f (x) = 2 x^{2} - x

x d x + sin^{2} (\frac{y}{x}) (y d x - x d y) = 0

\int (r^{3} + 1)^{2} \cdot r^{2} d r

\int \frac{x ^{4} - x ^{3}}{x ^{5}} d x

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