Integrals of square roots

[pointless132]

Senior in vector analysis

Im trying to figure out how to integrate square root of (1/4x^2 +3 + 9x^2) dx. (I mistakenly put 9t^2 previously but now is corrected).

I simply don't know how to even start. U-substitution won't work so I am just stuck. I have tried to search this forum for any similar problems but don't really see much that helps me out immediately.
I have done no work so I will need help from the beginning.

 

[Ishuda]

Senior in vector analysis

Im trying to figure out how to integrate square root of (1/4x^2 +3 + 9t^2) dx.

I simply don't know how to even start. U-substitution won't work so I am just stuck. I have tried to search this forum for any similar problems but don't really see much that helps me out immediately.
I have done no work so I will need help from the beginning.

Well WolframAlpha gives a solution (of sorts)
http://www.wolframalpha.com/input/?i=integrate+sqrt%28x^2%2Ba%29%2Fx+dx
maybe that will give some ideas.

 

[Deleted member 4993]

Senior in vector analysis

Im trying to figure out how to integrate square root of (1/4x^2 +3 + 9t^2) dx.

I simply don't know how to even start. U-substitution won't work so I am just stuck. I have tried to search this forum for any similar problems but don't really see much that helps me out immediately.
I have done no work so I will need help from the beginning.

Is it:

\(\displaystyle \displaystyle{\int \frac{1}{\sqrt{4x^2+3+9t^2}} \ dx}\)

or

\(\displaystyle \displaystyle{\int \left [\sqrt{\frac{1}{4}x^2+3+9t^2}\right]dx}\)

 

[pointless132]

Is it:

\(\displaystyle \displaystyle{\int \frac{1}{\sqrt{4x^2+3+9t^2}} \ dx}\)

or

\(\displaystyle \displaystyle{\int \left [\sqrt{\frac{1}{4}x^2+3+9t^2}\right]dx}\)

Its the second one except the x^2 is in the denominator with the 4 and i mistakenly put 9t^2. It should be 9x^2.
I tried to look for more info on this and i still have got nowhere.

 

[Steven G]

Its the second one except the x^2 is in the denominator with the 4 and i mistakenly put 9t^2. It should be 9x^2.
I tried to look for more info on this and i still have got nowhere.

Try multiplying/ and dividing by 4x^2 and then factor. Not writing it clearly really made this hard for everyone.

 

[Ishuda]

Its the second one except the x^2 is in the denominator with the 4 and i mistakenly put 9t^2. It should be 9x^2.
I tried to look for more info on this and i still have got nowhere.

I misunderstood and figured it was a function of two variables so that the 9 t² was a constant as far as the integral was concerned. So, start again
\(\displaystyle \sqrt{\frac{1}{4 x^2} + 3 + 9 x^2} = \sqrt{\frac{1 + 12 x^2 + 36 (x^2)^2}{4 x ^2}}\)
So first u = x^2; du/2x = du/(2u^1/2) = dx and we have
\(\displaystyle \sqrt{\frac{1}{4 x^2} + 3 + 9 x^2}\, \, dx = \sqrt{\frac{1 + 12 u + 36 u^2}{16 u^2}}\, du = \frac{1}{4} \frac{\sqrt{1 + 12 u + 36 u^2}}{u}\, du\)
\(\displaystyle = \frac{1}{4} \frac{\sqrt{(1 + 6u)^2}}{u}\, du\)
and you shoud be able to take it from there.

 

[pointless132]

I misunderstood and figured it was a function of two variables so that the 9 t² was a constant as far as the integral was concerned. So, start again
\(\displaystyle \sqrt{\frac{1}{4 x^2} + 3 + 9 x^2} = \sqrt{\frac{1 + 12 x^2 + 36 (x^2)^2}{4 x ^2}}\)
So first u = x^2; du/2x = du/(2u^1/2) = dx and we have
\(\displaystyle \sqrt{\frac{1}{4 x^2} + 3 + 9 x^2}\, \, dx = \sqrt{\frac{1 + 12 u + 36 u^2}{16 u^2}}\, du = \frac{1}{4} \frac{\sqrt{1 + 12 u + 36 u^2}}{u}\, du\)
\(\displaystyle = \frac{1}{4} \frac{\sqrt{(1 + 6u)^2}}{u}\, du\)
and you shoud be able to take it from there.

I got it! Thanks a lot!