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**delgeezee** Good News, I retried the example problem today and after a while I was able to get the answer. As you all suggested, I found that allowing$\displaystyle u=x+1$ to be the simplest approach. I also now understand why the the numerator changed to$\displaystyle u-1=x$. It was so, to keep the fuction the same, but in terms of u.

I am now attempting, with slight difficulty, to try the other substition example by letting $\displaystyle u= \sqrt{x+1}$

$\displaystyle \int{\frac{x}{\sqrt{x+1}} \dx$ = $\displaystyle \int{\frac{x}{u}dx$

Now I change the Numerator in terms of u.

$\displaystyle x= \sqrt{x+1} * \sqrt{x+1} - 1 \Rightarrow u*u-1 \Rightarrow u^2-1$

Now I need to change the variables for the dx in terms of du but I am having troubles.

my first inclination was that since$\displaystyle u = \sqrt{x+1}$, then du = $\displaystyle \frac{1}{2\sqrt{x+1}}$, then replace that x within the du to $\displaystyle \frac{1}{2\sqrt{(u^2-1)+1}}$ = $\displaystyle \frac{1}{2\sqrt{u^2}}$ = $\displaystyle \frac{1}{2u}$

So in the original function, $\displaystyle \int{\frac{x}{\sqrt{x+1}}dx$ , the dx= 1, so I should set the dx = du???

1 = $\displaystyle \frac{1}{2u}$ , multiply both sides to isolute du in terms of u, so finally, du=$\displaystyle 2u$

So Now I have $\displaystyle \int{\frac{u^2-1}{u}*2udu$ which is what the book also has

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In regards to finding the du, can I simply say Hey, the derivative of the numerator $\displaystyle u^2-1$ is 2u and i tack it on to the back of the new integrand to get $\displaystyle \int{\frac{u^2-1}{u}*2udu$ ???????

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Are these the correct logical steps I should take ?????