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

Question

Ezekiel Alexander · Accepted Answer

undefined To test the improper integral ∫(x-1)^-2+(x-3)^-2 dx from [1,3] and evaluate it, you first need to check for convergence. This integral is improper at both endpoints, as the integrand approaches infinity at x=1 and x=3.

To evaluate the integral, split it into two separate integrals at x=2, where the singularity occurs. Then, integrate each part separately from 1 to 2 and from 2 to 3.

The integral from 1 to 2 will have the form ∫(x-1)^-2 dx, and the integral from 2 to 3 will have the form ∫(x-3)^-2 dx.

After integrating each part, evaluate the limits as they approach 1 and 3, respectively. If both integrals converge, add their values together to find the total value of the improper integral. If either integral diverges, then the total integral is considered to diverge.

Christopher Agresta · Answer

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"#