What is the integral of (3x)/(2e^{6x)} ?

The integral of (3x)/(2e^{6x)} is 3/2 (-1/6 e^{-6x}x-1/36 e^{-6x})+C

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integral of (3x)/(2e^{6x)}

\int \frac{3 x}{2 e ^{6 x}} d x

\frac{3}{2} (- \frac{1}{6} e^{- 6 x} x - \frac{1}{36} e^{- 6 x}) + C

Solution steps

\int \frac{3 x}{2 e ^{6 x}} d x

Take the constant out:

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

= \frac{3}{2} \cdot \int \frac{x}{e ^{6 x}} d x

Apply Integration By Parts

: - \frac{1}{6} e^{- 6 x} x - \int - \frac{1}{6} e^{- 6 x} d x

= \frac{3}{2} (- \frac{1}{6} e^{- 6 x} x - \int - \frac{1}{6} e^{- 6 x} d x)

\int - \frac{1}{6} e^{- 6 x} d x = \frac{1}{36} e^{- 6 x}

= \frac{3}{2} (- \frac{1}{6} e^{- 6 x} x - \frac{1}{36} e^{- 6 x})

Add a constant to the solution

= \frac{3}{2} (- \frac{1}{6} e^{- 6 x} x - \frac{1}{36} e^{- 6 x}) + C

\int_{0}^{5} \frac{1}{x ^{\frac{15}{14}}} d x

\frac{d}{d x} (ln (x - 4 y))

f (x) = (x^{3} + 3) e^{x}

\frac{1}{6} x^{2} - 3 x + 15

\int cos (y) cos (4 y) d y

What is the integral of (3x)/(2e^{6x)} ?
The integral of (3x)/(2e^{6x)} is 3/2 (-1/6 e^{-6x}x-1/36 e^{-6x})+C