How do you integrate #int x^2e^(4x)# by integration by parts method?

2 Answers
Mar 21, 2017

The answer is #=e^(4x)(x^2/4-x/8+1/32)+C#

Explanation:

Integration by parts is

#intuv'dx=uv-intu'vdx#

Here, we have

#u=x^2#, #=>#, #u'=2x#

#v'=e^(4x)#, #=>#, #v=e^(4x)/4#

Therefore,

#intx^2e^(4x)dx=(x^2e^(4x))/4-int2x*e^(4x)dx/4#

#=(x^2e^(4x))/4-1/2intx*e^(4x)dx#

We apply the integration by parts a second time to find #intx*e^(4x)dx#

#u=x#, #=>#, #u'=1#

#v'=e^(4x)#, #=>#, #v=e^(4x)/4#

Therefore,

#intx*e^(4x)dx=x/4e^(4x)-inte^(4x)dx/4#

#=x/4e^(4x)-e^(4x)/16#

Putting the results together,

#intx^2e^(4x)dx=(x^2e^(4x))/4-1/2(x/4e^(4x)-e^(4x)/16)#

#=e^(4x)(x^2/4-x/8+1/32)+C#

Mar 30, 2017

There is a faster method

Explanation:

Introduce #t# and write
#I = int e^(4tx) dx#
Now differentiate with respect to t
#(dI)/dt = int 4x e^(4tx) dx#
Once more
#(d^2I)/dt^2 = int 16x^2 e^(4tx) dx#
#I = e^(4tx)/(4t)#
#(dI)/dt= x/te^(4tx) - e^(4tx)/(4t^2)#
and again
#(d^2 I)/(dt^2) = ( 4x^2/t -x/(t^2)-x/t^2 + 2/t^3 1/4)e^(4tx)#
#(d^2I)/dt^2 = (4x^2-2x + 2)e^(4x)#
Our integral is 1/16 I
#J= I/16 = (x^2 - x/4 + 1/8)e^(4x)#