Question

Integrate x/(x+1)....?

Answer 1

why not rewrite it as

\[\frac{x}{x + 1} = \frac{x + 1}{x + 1} - \frac{1}{x + 1}\]

then simplify it it and you get

\[\frac{x}{x + 1} = 1 - \frac{1}{x + 1}\]

now you should be able to integrate that.

hope it helps

Answer 2

1. The problem statement, all variables and given/known data
How do you integrate S x/(x-1) dx


2. Relevant equations

IBP

3. The attempt at a solution
I tried using Integration by Parts with u= x/(x-1) and dv = dx and that did not work out. Then I tried using u= 1/(x-1) and dv = xdx but that did not work either.

A google search said "Just use the substitution u = x+1, then replace dx with du and you get u+1/u = 1 + 1/u which you can integrate to give u + ln u, thus = x-1 + ln(x-1)". But I am pretty sure that you can't just add +1 to the numerator like that.

I also know that it can be done using tables but is there a way not to use tables? Did I do my integration wrong?

Thanks

Answer 3

@jordanloveangel why not? campell not just "add 1" , she did "-1" also.

Answer 4

campell's solution is perfect, to me.

Answer 5

I think @jordanloveangel just posted a solution he found elsewhere

Answer 6

@campbell_st 's method is the way I would do this.

Answer 7

Now since \(\dfrac{x}{x+1} \implies \dfrac{x+1}{x+1} - \dfrac{1}{x+1} \implies 1-\dfrac{1}{x+1}\) You have:

\[\int \left(1 -\frac{1}{x+1}\right)dx \implies \int 1dx -\int \dfrac{1}{x+1}dx\]This is when you let \(u = x+1 ~,~ du =dx\)

After subbing that in, you should know \(d(\ln(x)) = \dfrac{1}{x} ~,~ \therefore ~ \int \dfrac{1}{x}dx = ~?\)`