How to Correctly Integrate x e^(-3x) Using Integration by Parts?

[b0rsuk]

Homework Statement

$$\int x e^-3x dx$$

Homework Equations

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

Integration by substitution not allowed

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.

 

[cragar]

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

 

[SammyS]

Homework Statement

$$\int x e^{-3x} dx$$
...


$$=-\frac{1}{3} x e^{-3x} -\frac{1}{9} e^{-3x} + C$$

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

Why do you think this is incorrect ?