# Integral (x e^-3x dx)

1. Jan 10, 2012

### b0rsuk

1. The problem statement, all variables and given/known data
$$\int x e^-3x dx$$

2. Relevant equations

$$\int f(x)g'(x) = f(x)g(x) - \int f'(x) g(x)$$

Integration by substitution not allowed

3. The attempt at a solution
$$f(x) = x, f'(x) = 1, g'(x) = e^{-3x}, g(x) = \int e^{-3x} dx = -\frac{1}{3}e^{-3x}$$
$$\int x e^{-3x} dx = x(-\frac{1}{3})e^{-3x} - \int - \frac{1}{3} e^{-3x} dx =$$
$$= -\frac{1}{3} x e^{-3x} + \frac{1}{3} \int e^{-3x} dx =-\frac{1}{3} x e^{-3x} -\frac{1}{9} e^{-3x} + C$$

Which is incorrect. I'm not sure how to integrate e^(-3x) properly.

2. Jan 10, 2012

### cragar

so you tried integration by parts. what is the derivative of e^(-3x). And then how would i integrate this.

3. Jan 10, 2012

### SammyS

Staff Emeritus
Why do you think this is incorrect ?