let x = 2tanθ
x^2+4 = 4tan^2θ+4 = 4sec^2θ
dx = 2sec^2θ dθ
now the integrand becomes
2sec^2θ/4sec^2θ dθ = 1/2 dθ
Now recall that tan(π/2) = ∞
This is just a reminder of one of the standard forms:
∫ dx/(x^2+a^2) = 1/a arctan(x/a)
Find out what kind of improper integral is given below
§dx/{x^4+4}....?,hence evaluate the integral with
upper boundary=infinity
lower boundary =0
i need help plz show me full work
4 answers
so what do i do from here now sir
Oh. I'm sorry. I misread the integral. x^4+4 is a bit trickier.
But, as usual, google turned up several discussions. You might care to start here:
http://www.freemathhelp.com/forum/threads/55678-difficult-integration-int-1-(1-x-4)-dx
But, as usual, google turned up several discussions. You might care to start here:
http://www.freemathhelp.com/forum/threads/55678-difficult-integration-int-1-(1-x-4)-dx
If you follow the threads of that web site, you will probably see how wolframalpha came up with its soution:
http://www.wolframalpha.com/input/?i=%E2%88%AB%5B0,infinity%5D+dx%2F(x%5E4%2B4)
http://www.wolframalpha.com/input/?i=%E2%88%AB%5B0,infinity%5D+dx%2F(x%5E4%2B4)