How to approach this integrand: [math]\frac{1}{ \sqrt{x} (x+1)}[/math]

[Al-Layth]

The integrand is:
[math]\frac{1}{ \sqrt{x} (x+1)}[/math]
so far I have thought of one substitution: u=sqrt(x) which works out well and massages the integrand into a familiar form that has standardised substitutions.

but I am trying to think of other ways to approach the problem and cannot think of anything else except just that one way.

what else could work?
thanks

 

[pka]

The integrand is:
[math]\frac{1}{ \sqrt{x} (x+1)}[/math]
so far I have thought of one substitution: u=sqrt(x) which works out well and massages the integrand into a familiar form that has standardised substitutions.

but I am trying to think of other ways to approach the problem and cannot think of anything else except just that one way.

what else could work?
thanks

What is the derivative of [imath]2\arctan(\sqrt{x})[/imath]?

 

[Steven G]

You can try u^2=x+1=> sqrt(u^2 - 1) = sqrt(x). Then try some trig sub to see where that will get you.

 

[Al-Layth]

What is the derivative of [imath]2\arctan(\sqrt{x})[/imath]?

well the expression u mentioned is the area function of my expression but how does this help

You can try u^2=x+1=> sqrt(u^2 - 1) = sqrt(x). Then try some trig sub to see where that will get you.

thanks
How did you come up with these substitutions.
thought process plz

also aside from substitutions
is there anything else>

 

[Steven G]

well the expression u mentioned is the area function of my expression but how does this help


thanks
How did you come up with these substitutions.
thought process plz

also aside from substitutions
is there anything else>

If u^2 = x+1, then sqrt(u^2 - 1) = sqrt(x). That is we have a way to express both sqrt(x) and x+1 in terms of u. The real questions are how do you express dx in terms of u and will this substitution help.
Please post back showing your work as I would like to see how well my hint worked.

 

[lookagain]

The integrand is:
[math]\frac{1}{ \sqrt{x} (x+1)}[/math]

Don't forget that there is a dx as a part of the complete problem. Note that there is only one square root term.
A recommendation is to let \(\displaystyle \ u = \sqrt{x}.\) Then, \(\displaystyle \ u^2 = x. \ \ So, \ \ (x + 1) = (u^2 + 1), \ \ \) and \(\displaystyle \ \ dx = 2u \ du. \)
You eliminate the square root term as compared to the previous couple of posts, and look for the simplification
of the fraction.

Please continue from there and please show your work in this thread.