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$$x \over x+1 $$

I'm actually trying to integrate the above function, but I was told to first use long division to simplify it, which I'm not really sure how to do.

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  • $\begingroup$ Regarding long division: would the result not be 1 and -1 is the reaminder. Or do you have to go further? $\endgroup$ – zoli Apr 30 '15 at 18:00
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Pretend you were trying to divide $(x+1)$ into $x$. Your first step would be to say that the leading part of $(x+1)$ divides the leading part of $x$ one time, leaving a remainder of $-1$. Then the leading part of $(x+1)$ goes into the leading part of $-1$ zero times so you are finished, and the answer is $1$ with a remainder of $-1$. Thus $$ \int \frac{x}{x+1}dx = \int \left[ 1 + \frac{-1}{x+1} \right] dx = \int 1 \,dx - \int \frac{1}{x+1} dx = x - \ln(x+1) + C $$

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  • $\begingroup$ Ah ok, so if I can't divide x+1 into -1, I just put -1 over x+1 like: -1/x+1. But if it was say another constant, like 10, would it be 10/x+1 ? $\endgroup$ – Jay Apr 30 '15 at 18:12
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An easier way is to note that $$\frac{x}{x+1} = \frac{x+1-1}{x+1}=\frac{x+1}{x+1}+\frac{-1}{x+1}=1-\frac{1}{x+1}.$$

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I know, this isn't exactly long division, but this works too, in terms of setting up for the integration:$$\int \frac x{x+1} \, dx= \int \frac{x+1-1}{x+1} \, dx=\int \frac{x+1}{x+1}-\frac 1{x+1} \, dx=\int1 -\frac 1{x+1} \, dx$$

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    $\begingroup$ The denominator is $x+1$. You then change it to $x-1$. Please edit accordingly. $\endgroup$ – Jordan Glen Apr 30 '15 at 17:58
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if you are trying to integrate $\frac{x}{x+1},$ you don't need to do long division; can be done with a change of variable. change of variable is a useful technique, and in particular for functions of the from $f(ax+b).$

here is it works in this case. $$u = x + 1, x = u-1$$ so $$\int \frac{x}{x+1}\, dx = \int\frac{u-1}{u}\, du = \int\left(1 - \frac 1u\right)\, du = u - \ln u+c = x - \ln|x-1| + c$$

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