# `int_0^2 1/(x-1)^2 dx` Explain why the integral is improper and determine whether it diverges or converges. Evaluate the integral if it converges

## Expert Answers

Integral is improper if we have to take limit in order to calculate it. This can happen if we have infinite values of integration or if the interval if integration contains point(s) where the function is not defined. The latter is the case here because the function we are integrating is not defined for `x=1.`

Because this point is within the interior of the interval of integration (not at the endpoint) we first must write this integral as a sum of two integrals.

`int_0^2 1/(x-1)^2 dx=int_0^1 1/(x-1)^2dx+int_1^2 1/(x-1)^2dx=`

Substitute `u=x-1` `=>` `du=dx.` Since `u=x-1` all bounds of integration become lower by 1.

`int_-1^0 1/u^2 du+int_0^1 1/u^2 du=-1/u|_-1^0-1/u|_0^1=`

`lim_(u to 0^-) -1/u+1/-1-1/1+lim_(u to 0^+) 1/u=`

Notice the use of directional limits (from the left for the first and from the right for the second integral).

`-(-infty)-2+infty=infty`

As we can see the integral diverges.

The image below shows the graph of the function and area under it corresponding to the value of the integral. Asymptotes of the graph are `x`-axis and line `x=1.` The other image (the red one) shows the graph of the function `f(u)=1/u.` There we see why the two directional limits have different values.

## See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Images:
Image (1 of 2)
Image (2 of 2)

Further Reading

Approved by eNotes Editorial