# Use reduction formulas in Exercise 64 to evaluate the integral:`intx^2e^(3x)dx` [Hint: First make a substitution] I think this is the reduction formula they are talking...

Use reduction formulas in Exercise 64 to evaluate the integral:

`intx^2e^(3x)dx`

[Hint: First make a substitution]

I think this is the reduction formula they are talking about:

`intx^(n)e^(x)dx=x^(n)e^x-nintx^(n-1)e^xdx`

I think the substitution that I have to make is making the e^(3x) to e^u, but then I get confused where to put du.

### 1 Answer | Add Yours

Yes, you are on the right track. You can use the substitution u = 3x, which then means that

du = 3dx (so, dx = du/3) and

`x^2 = (u/3)^2 = u^2/9`

When you plug all this into original integral, it will become

`int u^2/9 e^u (du)/3` = `1/27 int u^2 e^u du`

Now you can apply the reduction formula with n = 2, which comes from the integration by part:

`int u^2 e^u du = u^2e^u - 2int ue^u du` + C ( C here and everywhere below is an arbitrary constant)

Now, apply the reduction formula again to the second integral, with n = 1:

`int ue^u du = ue^u - int e^udu = ue^u - e^u` +C

Putting this back into the first reduction formula results in

`int u^2 e^udu = u^2e^u - 2(ue^u - e^u) = u^2e^u - 2ue^u + 2e^u` +C

Now, recall that the original integral (before the substitution) was 1/27 of the integral in terms of u. Substituting u = 3x back, we get

`1/27 ((3x)^2 e^(3x) - 2(3x)e^(3x) + 2e^(3x))` +C

Multiplying this out results in

`x^2/3 e^(3x) - 2/9e^(3x) + 2/27e^(3x)` + C.

**This is the final result for the given integral.**