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Find the surface area of revolution of `y = 1/x` from `xgt=1` The setup of the integral is SA = `2pi int_0^oo(1/x*sqrt(1+1/x^2))dx` Then, you need...

There seems to be a discrepancy in your question: if `x >=1` , then the limits of the integral should be 1 and infinity. This means it would be an improper integral and it would not converge (as you will see after we take the integral). I am going to take the indefinite integral - without the limits - and the appropriate limits can be plugged in later.


Before taking the integral, let's simplify the radical expression under the radical:


`sqrt(1 +1/x^2) = sqrt((x^2+1)/x^2) = sqrt(1+x^2)/x` .


The integral then becomes (omitting the coefficient of `2pi` for now):


`int sqrt(1+x^2)/x^2 dx` . Notice that it contains the radical expression `sqrt(1 + x^2)` . The traditional way to take such integrals is by using trigonometric substitution


`x = tant` . Then, `dx = sec^2tdt` and `t = arctanx` .


The advantage of this substitution is that it allows to simplify the radical expression using the trigonometric identity `tan^2t + 1 = sec^2t` .


`sqrt(1+x^2) = sqrt(1+tan^2t) = sqrt(sec^2t) = sect` .


Plugging the above into the integral, we get


`int (sect)/(tan^2t) sec^2tdt` . This can be rewritten again using the same trigonometric identity as


`int (sect)/(tan^2t) (tan^2t + 1)dt = int sectdt + int (sect)/(tan^2t) dt` .


The first integral equals


`int sect dt = ln|sect + tant|` . Substituting the original variable x back, we get `ln|sqrt(1+x^2) +x|` .


To take the second integral, use `sect = 1/(cost)` and `tan^2t = (sin^2t)/(cos^2t)` :


`int (1/(cost))/((sin^2t)/(cos^2t)) dt = int (cost)/(sin^2t) dt` . Here, we can use substitution


u = sint. Then, du = costdt and the integral becomes


`int (du)/u^2 = -1/u = -1/(sint)` . In terms of x = tant,


`sint = tantcost = (tant)/(sect) = x/sqrt(1+x^2)` .


Putting it all together, we see that the original integral (without the limits) equals to


`2piln|sqrt(1+x^2) + x| - 2pisqrt(1+x^2)/x` + C (C is an arbitrary constant.)

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