integration qs (1 Viewer)

[*Pooja*]

can sum1 help me with these:

1) (1+x)/root of (1-x-x^2) dx

2) x/(root of x + 1) dx

thanks.

 

[tina_goes_doo]

Ok i suck at these but i'll have a shot at the second one:

x/(root of x + 1) dx

let u = x + 1
x = u - 1
du = dx

:. x/(sqrt of x + 1) dx
= (u - 1) / (sqrt u) du

Break these into two integrals:

= u / (sqrt u) du - 1 / (sqrt u)

= u^(1/2) du - u^(-1/2)

and from there it should be a simple integation.

 

[+Po1ntDeXt3r+]

1) (1+x)/root of (1-x-x^2) dx

break this up..

I=S (x+1/2)/root of (1-x-x^2) + 1/(2*root of (1-x-x^2)) dx

first part chain rule second inverse trig
where
1/(2*root of (1-x-x^2)) = 1/(2*root of (5/4 - (x+1/2)^2)

cheers

soln: highlight for spoiler

I = 1/2*arcsin(2/5*5^(1/2)*(x+1/2))-(1-x-x^2)^(1/2) + C
n.b: arcsin means inverse function of sine

 

[*Pooja*]

oops sorry u took me wrong. number 2) is actually x/[(root of x)+1]. sorry about that. my mistake...

 

[tina_goes_doo]

Yeah sorry, thought that was a pretty alright question to do!

 

[CM_Tutor]

Originally posted by *Pooja*
x/[(root of x)+1]

Do a substitution, u² = x

 

[KeypadSDM]

Substitution? That's a bit silly, and slow.

Clever manipulation does the job perfectly:

x/(1 + Sqrt[x])
= Sqrt[x] - Sqrt[x]/(1 + Sqrt[x])
= Sqrt[x] - 1 + 1/(1 + Sqrt[x])
= Sqrt[x] - 1 + 1/(Sqrt[x]) + 2((1/(2Sqrt[x]))/(Sqrt[x] + 1))
Integrating gives:

2x^3/2/3 - x + 2Sqrt[x] + 2Ln[Sqrt[x] + 1] + C

It's a nice function to work with.