How would one solve the following equation? This equation is giving me a hard time.
$$e^x(x^2+2x+1)=2$$
Can you show me how to solve this problem algebraically or exactly? I managed to solve it using my calculator with one of its graph functions. But I would like to know how one would solve this without using the calculator.
Highly appreciated,
Bowser.
 A: Take the natural logarithm of both terms, getting
$$x + \ln(x^2 + 2x + 1) = \ln(2)$$
Now watch the log argument: you see it's a square! Indeed $x^2 + 2x + 2 = (x + 1)^2$ so, again using log property:
$$x + 2\ln(x+1) = \ln(2)$$
If we assume to expect small values as a solution, then we may use Taylor Series expansion for the logarithm, up to the second order:
$$\ln(1+x) \approx x - \frac{x^2}{2}$$
Thence:
$$x + 2x - x^2 - \ln(2) = 0 ~~~~~ \to ~~~~~ x^2 - 3x - \ln(2) = 0$$
Solving like a second degree equation gives
$$x = \frac{3\pm \sqrt{9 - 4\ln(2)}}{2}$$
$$x_1 = 2.747(..) ~~~~~~~~~~~ x_2 = 0.252(..)$$
This is a numerical method to solve it and as you see the second solution fits with your result.
That solution can be improved simple taking more terms in the log expansion, indeed:
$$\ln(x+1) \approx x - \frac{x^2}{2} + \frac{x^3}{3} \cdot $$
A: The answer given by Desmos for intersection of the two curves $y=e^x$ and $y=\frac {2}{(x+1)^2}$ is  $\color{red}{x=0.249}$. Now we have
$$(x+1)^2=2e^{-1}\iff x^2+2x+1=2(1-x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{24}-\frac{x^5}{60}+O(x^6))$$ hence $$1-4x-\frac{x^3}{3}+\frac{x^4}{12}-\frac{x^5}{60}+20\cdot O(x^6)=0$$
The first approximation $1-4x=0$ gives $\color{red}{x\approx 0.25}$
The second approximation $ 1-4x-\frac{x^3}{3}=0$ gives $\color{red}{x\approx 0.24872}$
And we can  continue but we see that the first approach is already good enough.
A: $$
e^x(x+1)^2=2\implies e^{(x+1)/2}(x+1)/2=\sqrt{e/2}\implies(x+1)/2=\mathrm{W}\!\left(\sqrt{e/2}\right)
$$
Therefore,
$$
x=2\mathrm{W}\!\left(\sqrt{e/2}\right)-1
$$
where $\mathrm{W}$ is the Lambert W function.
There is an iterative algorithm given in this answer to compute $\mathrm{W}$.
Alternatively, N[2LambertW[Sqrt[E/2]]-1,20] in Mathematica yields 
$$
x=0.24879269668640244047
$$
