Integral - Algebraic Manipulation?

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Homework Statement



\int \frac{x}{x+d}dx

The Attempt at a Solution



I tried parts and that turned out horribly so I fired up maple. First move it makes is 'Rewrite Rule'. I've never heard of a rewrite rule but I supposed it could be some algebraic massaging. I ended up looking at it this way:

\frac{x}{x+d} = 1-1+\frac{x}{x+d} = \frac{x+d}{x+d}-\frac{x+d}{x+d}+\frac{x}{x+d} = \frac{x+d}{x+d}+\frac{x-d-x}{x+d} = 1-\frac{d}{x+d}

Which then is easily integrable. My question is, is there a snappier way to think of this/do it?
 
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Just divide x by x+d and you'll get the term on the rightmost of your equalities.
 
As in polynomial long division?

edit: Or is there some way that is easy to see what will happen? Basically it isn't straight up intuitive for me to realize that x/x+d is 1-d/x+d, I'm trying to find a quick method to deal with it.
 
Last edited:
\int\frac{x+d-d}{x+d}dx

Break it and integrate.
 
That's much shorter than what I did, thanks.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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