How do you integrate #x^2 e^-5x dx#?

Question

Samantha Adkison · Accepted Answer

Hello, I think you want integrate #x^2 e^{-5x}#, not #x^2 e^{-5}x#... Use two integrations by parts :
#\int_0^x u(t) v'(t) dt = [u(t)v(t)]_0^x - \int_0^x u'(t) v(t) dt#.
You can prove that if you write #(uv)' = u'v+uv'#  and if you integrate between #0# and #x#.  Let #F# be the function define by #F(x) = \int_0^x t^2 e^{-5t} dt#. Take #u(t) = t^2# and #v(t) =-\frac{1}{5}e^{-5t}#. So you have 
#u'(t) = 2t# and #v'(t) = e^{-5t}# Utilize the formula for integration by parts: #F(x) = [-\frac{1}{5} t^2 e^{-5t} ]_0^x +\frac{2}{5} \int_0^x te^{-5t}dt# Use integration by parts again to calculate #\int_0^x te^{-5t}dt#. Take
#u(t) =t# and #v(t) = -\frac{1}{5}e^{-5t}#. So you have
#u'(t) = 1# and #v'(t) = e^{-5t}#, and  #\int_0^x te^{-5t}dt = [-\frac{1}{5}t e^{-5t}]_0^x + \frac{1}{5}\int_0^x e^{-5t}dt =  [-\frac{1}{5}t]_0^x + \frac{1}{5}[-\frac{1}{5}e^{-5t}]_0^x# You could make things simpler: #\int_0^x te^{-5t}dt =-\frac{1}{5}xe^{-5x} - \frac{1}{25}(e^{-5x}-1)# Lastly, enter that into the expression above: #F(x) = -\frac{1}{5}x^2e^{-5x} - \frac{2}{5}(\frac{1}{5}xe^{-5x} + \frac{1}{25}(e^{-5x}-1))# Primitives (or antiderivatives) of #x^2e^{-5x}# are  #-(\frac{1}{5}x^2 + \frac{2}{25}x + 2/125)e^{-5x} + c# where #c \in RR#.

Isaiah Allen · Answer

undefined To integrate $ x^2 e^{-5x} \, dx $, you can use integration by parts. Let $ u = x^2 $ and $ dv = e^{-5x} \, dx $. Then, $ du = 2x \, dx $ and $ v = -\frac{1}{5} e^{-5x} $.

Now, applying the integration by parts formula:

$$ \int u \, dv = uv - \int v \, du $$

We have:

$$ \int x^2 e^{-5x} \, dx = -\frac{x^2}{5} e^{-5x} - \int -\frac{2x}{5} e^{-5x} \, dx $$

Integrating the second term again by parts:

Let $ u = x $ and $ dv = e^{-5x} \, dx $. Then, $ du = dx $ and $ v = -\frac{1}{5} e^{-5x} $.

$$ -\frac{x^2}{5} e^{-5x} - \left( -\frac{x}{5} e^{-5x} - \int -\frac{1}{5} e^{-5x} \, dx \right) $$

$$ -\frac{x^2}{5} e^{-5x} + \frac{x}{5} e^{-5x} - \frac{1}{25} e^{-5x} + C $$

So, the integral of $ x^2 e^{-5x} \, dx $ is $ -\frac{x^2}{5} e^{-5x} + \frac{x}{5} e^{-5x} - \frac{1}{25} e^{-5x} + C $, where $ C $ is the constant of integration.