How do you integrate #(x^5)*(e^(x^2/2))#?

1 Answer
Gaurav · mason m
Sep 1, 2015

#(4x^4-4x^2-8)e^(x^2/2)+c#

Explanation:

Let's assume #x^2/2 = t#
#=> (2xdx)/2 = t, x dx = dt#

#=int(x^5)*(e^(x^2/2))dx# can be written as,

#=int(x^2)(x^2)x(e^(x^2/2))dx#

Substituting with #t# we get

#=int(2t)(2t)(e^t)dt#

#=4intt^2(e^t)dt#

Using Integration by Parts,

#∫(I)(II)dx=(I)∫(II)dx−∫((I)'∫(II)dx)dx#

where #(I)# and #(II)# are functions of #x#, and #(I)# represents which will be differentiated and #(II)# will be integrated subsequently in the above formula

Similarly following for the problem,

#=4(t^2)inte^tdt-4int((t^2)'inte^tdt)dt+c#

#=4t^2e^t-4int2te^tdt+c#

#=4t^2e^t-8intte^tdt+c#

Applying again integration by parts in second term, we get

#=4t^2e^t-8(te^t-inte^tdt)+c#

#=4t^2e^t-8(te^t-e^t)+c#

#=4t^2e^t-8te^t-8e^t+c#

Substituting #t# back,

#=4(x^2/2)^2e^(x^2/2)-8(x^2/2)e^(x^2/2)-8e^(x^2/2)+c#, where #c# is a constant

#=(4x^4-4x^2-8)e^(x^2/2)+c#

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