Question

Determine whether the integral converges.

I have some work, but I'm not sure it's correct. Help?

Answer 1

You have one mistake which occurs when you integrate
`\frac1{1-p^2}`. The antiderivative of this is not in terms of
`\tan^(-1)p`. Instead, letting
`p=\sin r` (or
`\cos r`, if you want to bother with more signs) gives
`\mathrm dp=\cos r\,\mathrm dr`, making the indefinite integral equality


`\displaystyle-\frac12\int(\mathrm dp)/(1-p^2)=-\frac12\int(\cos r)/(1-\sin^2r)\,\mathrm dr=-\frac12\int\sec r\,\mathrm dr=\ln|\sec r+\tan r|+C`

and then compute the definite integral from there.


`-\frac12\ln|\sec r+\tan r|\stackrel{r=\sin^(-1)p}=-\frac12\ln\left|(1+p)/(√(1-p^2))=\ln\left|\sqrt{(1+p)/(1-p)}\right|`

`\stackrel{p=u/2}=-\frac12\ln\left|\sqrt{(1+\frac u2)/(1-\frac u2)}\right|=-\frac12\ln\left|\sqrt{(2+u)/(2-u)}\right|`

`\stackrel{u=x+1}=-\frac12\ln\left|\sqrt{(3+x)/(1-x)}\right|`

`\implies-\frac12\displaystyle\lim_(t\to\infty)\ln\left|\sqrt{(3+x)/(1-x)}\right|\bigg|_(x=2)^(x=t)=-\frac12\left(\ln|-1|-\ln\left|\sqrt{\frac5{-1}}\right|\right)=\frac{\ln\sqrt5}2=\frac{\ln5}4`

Or, starting from the beginning, you could also have found it slightly more convenient to combine the substitutions in one fell swoop by letting
`x+1=2\sec y`. Then
`\mathrm dx=2\sec y\tan y\,\mathrm dy`, and the integral becomes


`\displaystyle\int_2^\infty(\mathrm dx)/((x+1)^2-4)=\int_(\sec^(-1)(3/2))^(\pi/2)(2\sec y\tan y)/(4\sec^2y-4)\,\mathrm dy`

`\displaystyle=\frac12\int_(\sec^(-1)(3/2))^(\pi/2)\csc y\,\mathrm dy`

`\displaystyle=-\frac12\ln|\csc y+\cot y|\bigg|_(y=\sec^(-1)(3/2)}^(y=\pi/2)`

`\displaystyle=-\frac12\lim_(t\to\pi/2^-)\ln|\csc y+\cot y|\bigg|_(y=\sec^(-1)(3/2))^(y=t)`

`\displaystyle=-\frac12\left(\lim_(t\to\pi/2^-)\ln|\csc t+\cot t|-\ln\frac5{\sqrt5}\right)`

`=\frac{\ln\sqrt5}2-\frac\ln2`

`=\frac{\ln5}4`

Another way to do this is to notice that the integrand's denominator can be factorized.


`x^2+2x-3=(x+3)(x-1)`

So,


`\frac1{x^2+2x-3}=\frac1{(x+3)(x-1)}=\frac14\left(\frac1{x-1}-\frac1{x+3}\right)`

There are no discontinuities to worry about since you're integrate over
`[2,\infty)`, so you can proceed with integrating straightaway.


`\displaystyle\int_2^\infty(\mathrm dx)/(x^2+2x-3)=\frac14\lim_(t\to\infty)\int_2^t\left(\frac1{x-1}-\frac1{x+3}\right)\,\mathrm dx`

`=\displaystyle\frac14\lim_(t\to\infty)(\ln|x-1|-\ln|x+3|)\bigg|_(x=2)^(x=t)`

`=\displaystyle\frac14\lim_(t\to\infty)\ln\left|(x-1)/(x+3)\right|\bigg|_(x=2)^(x=t)`

`=\displaystyle\frac14\left(\lim_(t\to\infty)\ln\left|(t-1)/(t+3)\right|-\ln\frac15\right)`

`=\displaystyle\frac14\left(\ln1-\ln\frac15\right)`

`=-\frac14\ln\frac15=\frac{\ln5}4`

Just goes to show there's often more than one way to skin a cat...