Is There a Simpler Method for Integrating xe^0.1x?

[morson]

Homework Statement

\int {x}{e^{0.1x}} dx

Homework Equations

U-substitution, differentiating.

The Attempt at a Solution

We have \int {x}{e^{0.1x}} dx

Let u = 0.1x therefore du = 0.1 dx ==> dx = 10du

Substituting back into the equation and using the fact that x = 10u:

\int {10u}{e^{u}} 10 du = 100 \int {u}{e^u} du

At this point I'm stuck. Is there another, simpler method?

 

[theperthvan]

Use integration by parts.

Let
u=x
dv=e^(0.1x) dx

 

[George Jones]

This looks OK.

To proceed further, what is another (besides substitution) technique of integration?

 

[morson]

This looks OK.

To proceed further, what is another (besides substitution) technique of integration?

I haven't been studying integration for very long, but I learned a method where you differentiate one part of the integral and express the integral in terms of the derivative, and then use the fact that \int f'(x) = f(x) + C

I'll have a shot:

From \int {x}{e^{0.1x}} dx

\frac {d}{dx} {x}{e^{0.1x}} = {e^{0.1x}} + {0.1}{x}{e^{0.1x}} = {e^{0.1x}}({1} + {0.1}{x})

I don't know how to express the integral in terms of f'(x), though.

 

[AlephZero]

You have found
\frac {d}{dx} xe^{0.1x} = e^{0.1x}(1 + 0.1x)

Multiply that by 10 and integrate both sides:

10x{e^{0.1x}} = \int 10 e^{0.1x}dx + \int xe^{0.1x} dx

You know how to integrate

\int 10 e^{0.1x}dx

This integration technique amounts to making a guess at what type of function the answer will be, and (if you guess right) reducing the problem to a simpler one.

For this problem the standard method of integration by parts (which you might not have learned yet) will produce the answer without the need to guess what form it might take.