Question

Use an appropriate change of variable to evaluate the convergent improper integral

1/[(x^0.5)(1+x)]

what should i use?

Answer 1

the answer is (pi) but i dunno how should i get to the answer

Answer 2

Is it dy/dx = 1/[(x^0.5)(1+x)] ? Move the dx to the right side, integrate the left and you get "y", integrate the right side now and try using a trig substitution...

Answer 3

sorry..my bad... 1st time use this website...but i found this fun ^ ^

the question is based on improper integral.
\[\int\limits\limits_{0}^{\infty} 1\div [ \sqrt{x} (1+x) ]\]

Answer 4

Okay, for the right hand side, you're going to use a u-substitution (try u = sqrt(x)) and after the sub, you should see a pretty clean inverse trig function that pops out.

Answer 5

Then, finding evaluating at the limits will be an easy task. :)

Answer 6

*eliminate "finding" :P typo.

Answer 7

thanks for helping but i cant get through with the trig part.. i cant see which i use to substitute i stuck at \[1\div (u + u ^{3})\] :(

Answer 8

I've been scratching my head at this one for a while, but I just got it. :) You've got du=1/(2sqrt(x)) dx, right? You've gotta multiply the left by 2sqrt(x) to get dx = 2√x du! Plug THAT into your integral and you'll get 2 * ∫ 1/(1+u^2) du. That should be much easier to solve!

Answer 9

hm...i getting 0 at the end .. hm... odd.. the answer i getting from my lecture is Pi what is the answer u getting? i get in the end \[\tan^{-1} u \]

Answer 10

Exactly, you're just forgetting to multiply by 2. So, your final expression in the end is 2 * arctan√x from 0 to ∞. The value of arctan(√x) as x approaches infinity is π/2, and as x approaches 0, is 0! So you get 2* (π/2-0) after applying the limits and the final value is π.

Answer 11

damn i keep doing so stupid mistake...i might need my brain rest for a while..XD..it's been 5 hours for me ... thanks..u r awesome.. ^ ^