Question

How would I solve the improper integral 1/(sqrt(x^2+1) from 0 to infinity

Answer 1

do you know the anti derivative of \(\frac{1}{\sqrt{x^2+1}}\)?

Answer 2

yes @satellite73 it's going to be arctan

Answer 3

what the hell is this???

Answer 4

what??

Answer 5

this business you have copied and pasted above that has nothing to do with this problem?

Answer 6

i miss my old delete button, need to get it back.

Answer 7

@Rohangrr last question also you copied and pasted from somewhere else!!!
Please use your own work here

Answer 8

@ash2326 i dont copy dam!! make that sure

Answer 9

I just know that improper integration of 1/(sqrt(x^2+1) is
\[\ln(x+\sqrt{x^2 + 1}) \]
Now apply upper limit - lower limit and get your answer

Answer 10

u dont know the ans... thats improper integral if i dont get to know i have ms word i solve in my copy n write there!! @ash2326

Answer 11

Ok then will you explain it @Rohangrr

Answer 12

@Rohangrr you are making it difficult to answer this question. the material you copied and pasted has nothing to do with the question asked

Answer 13

sorry i had written some thing else!! @satellite73 yep sorry dude

Answer 14

neither did your last 3 answers so maybe it is time for you to take a break

Answer 15

@shivam_bhalla that's what I got for the improper integral now all I have to do is evaluate at the limits thank you :)

Answer 16

lets repost and we can do this problem without much difficutly

Answer 17

:) Welcome

Answer 18

I will post my next question shortly I am doing exam prep and I just needed help with a couple series and improper integral questions thanks for your assistance

Answer 19

@satellite73 We can evaluate the indefinite integral but the limits here 0 to \(\infty\). I think the integral doesn't converge for these limits

Answer 20

@Rohangrr Wolfram much?

Answer 21

no it does not converge because as x gets large, so does \(\ln(x)\)

Answer 22

@Romero dont tungsten much!

Answer 23

@Rohangrr Did you just wolframed wolfram ?