Hi can someone please help me with this and explain how you get the answer.

find the integral of X/(sqrtX^2+4) dx

any help would be much appreciated.

Thanks

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- Sep 9th 2012, 11:54 PMrich1979struggling with integration
Hi can someone please help me with this and explain how you get the answer.

find the integral of X/(sqrtX^2+4) dx

any help would be much appreciated.

Thanks - Sep 9th 2012, 11:58 PMMarkFLRe: struggling with integration
I assume you are given:

$\displaystyle \int\frac{x}{\sqrt{x^2+4}}\,dx$

What do you think is a good candidate for a*u*-substitution? - Sep 10th 2012, 12:15 AMrich1979Re: struggling with integration
yeah am not really sure where to start with this one.

i thought that u=sqrtx^2+4 which makes it x+4 correct? - Sep 10th 2012, 12:27 AMMarkFLRe: struggling with integration
Isn't the entire expression $\displaystyle x^2+4$ under the radical?

If so, then try:

$\displaystyle u=x^2+4\:\therefore\:du=2x\,dx$

Is there a way for us to manipulate the integral so that we have a $\displaystyle 2x\,dx$ inside the integral? - Sep 10th 2012, 01:15 AMrich1979Re: struggling with integration
im sure there is but i cant figure it out at the min. must have missed the part of class where they explained this

- Sep 10th 2012, 01:20 AMMarkFLRe: struggling with integration
Multiply by 2 inside and by 1/2 in front, for a net result of multiplying by 1. So that you have:

$\displaystyle \frac{1}{2}\int\frac{2x}{\sqrt{x^2+4}}\,dx$

Now, using the aforementioned substitution, how may we write this? - Sep 10th 2012, 01:45 AMrich1979Re: struggling with integration
think i have it 1/2 int 1/sqrt U du

so answer once you substitute back is sqrt(X^2+4.)+C - Sep 10th 2012, 02:02 AMMarkFLRe: struggling with integration
Yes, that is correct. :)

- Sep 10th 2012, 02:13 AMrich1979Re: struggling with integration
thank you for your help and patience !!!

- Sep 10th 2012, 02:16 AMMarkFLRe: struggling with integration
Glad to help and glad you stuck it out! :)