The problem involves evaluating the integral ∫[√x /(1+x)] dx using substitution. The original poster has attempted a substitution with u = √x, leading to a new integral form that requires further simplification.
Some participants have provided guidance on using polynomial long division and an algebraic trick to simplify the integral. There is a shared concern regarding the correctness of the book's answer, with multiple participants expressing skepticism about the provided solution.
Participants note the need for algebraic steps to continue with the integration process and question the validity of the book's answer, which does not align with their findings.
This looks OK. I would use polynomial long division to turn this improper rational expression into a nicer form for integration. If you're not familiar with this technique, do a web search for "polynomial long division".ArmandStarks said:Homework Statement
The problem is this:
∫[√x /(1+x)] dx
I used sustitution method
u= √x
u^2=x
2udu=dxHomework Equations
My new integral is:
∫[(u*2udu)/(1+u^2)]du
2∫[u^2/(1+u^2)]du
I agree - the book's answer doesn't look right.ArmandStarks said:I need help to reduce this point to continue, I guess I need some algebraic steps
The Attempt at a Solution
The book shows: x^3/3 -3x - 10Ln|x-5| + C as the answer but I think this is wrong because I don't have x-5 anywhere
Hello ArmandStarks. Welcome to PF !ArmandStarks said:Homework Statement
The problem is this:
∫[√x /(1+x)] dx
I used sustitution method
u= √x
u^2=x
2udu=dx
Homework Equations
The Attempt at a Solution
My new integral is:
∫[(u*2udu)/(1+u^2)]du
2∫[u^2/(1+u^2)]du
I need help to reduce this point to continue, I guess I need some algebraic steps
The book shows: x^3/3 -3x - 10Ln|x-5| + C as the answer but I think this is wrong because I don't have x-5 anywhere
I just found that "trick" and the 2 integrals are easy to do.SammyS said:Hello ArmandStarks. Welcome to PF !
You can use "long division" to divide u2 by (u2 + 1) .
Or use the following "trick" .
##\displaystyle \frac{u^2}{1+u^2}=\frac{u^2+1-1}{1+u^2}##
##\displaystyle =\frac{1+u^2}{1+u^2}-\frac{1}{1+u^2}##