How do you test the improper integral #int (x-1)^-2+(x-3)^-2 dx# from #[1,3]# and evaluate if possible?

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Answer 1 · 2018-06-08T12:01:31

The improper integral diverges

Compute the indefinite integral

#int(1/(x-1)^2+1/(x-3)^2)dx=-1/(x-1)-1/(x-3)+C#

Compute the boundaries

#lim_(x->1^+)(-1/(x-1)-1/(x-3))=lim_(x->1^+)(-1/(x-1))-lim_(x->1^+)(-1/(x-3))#

#=-oo+1/2#

#=-oo#

#lim_(x->3^-)(-1/(x-1)-1/(x-3))=lim_(x->3^-)(-1/(x-1))-lim_(x->3^-)(-1/(x-3))#

#=-1/2-(-oo)#

#=+oo#

Finally,

#int_1^3(1/(x-1)^2+1/(x-3)^2)dx= " diverges"#