Is someone able to show me how I'd calculate this integral (showing working steps)? Any help would be greatly appreciate.
$\displaystyle \int{{-5x^3 e^{-5x}}\over{3}} - {5x^2 e^{-5x} dx}$
This is the same as
$\displaystyle \tfrac{1}{3}\int\left[-5x^3e^{-5x}+3x^2e^{-5x}\right]\,dx-\int6x^2e^{-5x}\,dx=\tfrac{1}{3}\int\frac{\,d}{\,dx}\left[x^3e^{-5x}\right]\,dx-6\int x^2e^{-5x}\,dx$ $\displaystyle =\tfrac{1}{3}x^3e^{-5x}-6\int x^2e^{-5x}\,dx$.
Now apply integration by parts twice to $\displaystyle \int x^2e^{-5x}\,dx$ and then you're done!
Can you take it from here?
Actually, something I'm confused about. Why did you bother making the equation:
$\displaystyle
\tfrac{1}{3}\int\left[-5x^3e^{-5x}+3x^2e^{-5x}\right]\,dx-\int6x^2e^{-5x}\,dx
$
rather than just:
$\displaystyle
\tfrac{1}{3}\int-5x^3e^{-5x}\,dx-\int5x^2e^{-5x}\,dx
$
Wouldn't the second option be equal, but easier to solve?
Let's see!
Let's focus on $\displaystyle \int -5x^3e^{-5x}\,dx$.
Let $\displaystyle u=-5x^3$ and $\displaystyle \,dv=e^{-5x}\,dx$
Then $\displaystyle \,du=-15x^2\,dx$ and $\displaystyle v=-\tfrac{1}{5}e^{-5x}$
Thus, $\displaystyle \int -5x^3e^{-5x}\,dx=x^3e^{-5x}-3\int x^2e^{-5x}\,dx$.
Now, note that $\displaystyle \tfrac{1}{3}\int -5x^3e^{-5x}\,dx-5\int x^2e^{-5x}\,dx=\tfrac{1}{3}\left[x^3e^{-5x}-3\int x^2e^{-5x}\,dx\right]-5\int x^2e^{-5x}\,dx$ $\displaystyle =\tfrac{1}{3}x^3e^{-5x}-\int x^2e^{-5x}\,dx-5\int x^2e^{-5x}\,dx=\tfrac{1}{3}x^3e^{-5x}-6\int x^2e^{-5x}\,dx$
So you'll end up with the same thing!
I just thought forcing product rule in the integrand would save time in integrating ==> you only need to apply integration by parts twice.