How do you find the antiderivative of #x^2 e^(2x)#?

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Answer 1 · 2018-06-03T13:41:50

#1/2e^(2x)(x^2 - x + 1/2) + C#

Well, This can't be done simply.

We have to Perform Integration By Parts.

The Rule States That,

#int uvdx = uintvdx - int (u'intv)dx#

Where #u# and #v# are functions of #x# and #u' = d/dx(u)#.

Now, We have,

#intx^2e^(2x)dx#

#= x^2inte^2x - int(d/dx(x^2)inte^(2x))dx#

#= 1/2x^2e^(2x) - int(cancel(2)x xx 1/cancel(2)e^(2x))dx#

#= 1/2x^2e^(2x) - int xe^(2x) dx#......................(i)

We have to Integrate By Parts Once More.

So, From (i),

#1/2x^2e^(2x) - x inte^(2x) + int(d/dx(x)inte^(2x))dx#

#= 1/2x^2e^(2x) - 1/2xe^(2x) + 1/2inte^(2x)dx#

#= 1/2x^2e^(2x) - 1/2xe^(2x) + 1/4e^(2x) + C# [Yeah, Don't EVER Forget this guy here#rarr# #C#]

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

Hope this helps.