Can you help me to solve this? thanks for help

Question

#int(1/x^2)(1/e^x)dx#

846 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-05T10:13:07

#-1/e^x+e^[[1-x]] -1/x#

Let #1/[x^2e^x]=A/e^x+B/x^2#, multiply denominator and numerator by #[x^2e^x]#. [Method of partial fractions]

So, #1= A[x^2e^x]/e^x# +#B[x^2e^x]/x^2#= #Ax^2 +Be^x#

Let #x=0#, therefore, #1= A[0]+Be^0#, ie, #[B=1]# ............[*]

Let #x= 1#, therefore, #1#= #A[1]#+#Be#,i.e, #A= [1-e]#,[since #B=1#

So, #int1/[x^2e^x]dx#=#int [[1-e]/e^x#+ #1/x^2]dx#=#int [1/ex -e/e^x+1/x^2]dx#,

These terms can now be integrated separately, giving the answer above. Hope this was helpful.