Graphing Y=1/x + 1/(x+1): Problem with Curve Sketching Methods

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in try to use curve sketching methods to graph this equation I've come across a problem in

y''=[2(x+1)^3+2x^3]/x^3(x+1)^3

I have verified this equation for y'' is correct but solving the numerator for critical numbers by hand ?
 
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How accurate does the graph need to be?

I mean you know how the graph 1/x looks like (I hope you do), 1/(x+1) is similar but with an asymptote in x=-1, the graph should be a superposition of these graphs, so if x+1>0, the graph of 1/(x+1) will tend more rapidly to zero than 1/x so it's dominant in the region of x>0, so the tendency of y to zero should be similar to 1/(x+1) from -1 to zero, and for 0<x<<1 the 1/x is dominant.
 
There several ways to solve x^3+(x+1)^3=0 by hand. i) Multiply it out and factor it. ii) There's a formula to factor the sum of two cubes. Or iii) write it as -x^3=(x+1)^3 and take the cube root. You'll lose the complex roots if you do it that way, but you don't need them for this problem anyway.
 
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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