Non-linear second order ODE I have to solve $$ y''(x)+(y'(x))^2=y'(x). $$ 
Using $ y'(x)=z $, I can write $$\int \frac{1}{z-z^2}dz=\int dx $$
So: 
$$\frac{1}{z(1-z)}=\frac{A}{z}+\frac{B}{1-z}$$
leads to
$$ \int \frac{1}{z(1-z)}dz=\int \frac{1}{z}dz+\int \frac{1}{1-z}dz= \ln(z)-\ln(1-z)$$
$$\Rightarrow \ln\left(\frac{z}{1-z}\right)=x+c $$
$$\Rightarrow z=\frac{e^{x+c}}{1+e^{x+c}}=y'$$
$$\Rightarrow y=\int \frac{e^{x+c}}{1+e^{x+c}}dx $$
Now, calling $e^{x+c}=t$:
$$y=\int\frac{t}{1+t}\cdot\frac{dt}{t}=\ln(1+t)\Rightarrow \ln(1+e^{x+c})$$
I checked the calculations and i thought was right but WolframAlpha says that the result is $\ln(c_1+e^x)+c_2$. What am I doing wrong? 
Thanks for any help!
 A: You didn't do anything wrong. The answer you got is $\ln(1+e^{x+c_1})+c_2$. It is equal to:
\begin{align}
\ln(1+e^{x+c_1})+c_2 &= \ln(e^{c_1}(e^x+\frac{1}{e^{c_1}}))+c_2 \\
&= \ln(e^{c_1})+\ln(e^x+\frac{1}{e^{c_1}})+c_2 \\
&= \ln(\frac{1}{e^{c_1}}+e^x)+(c_1+c_2). 
\end{align}
Now let $d_1=\frac{1}{e^{c_1}}$, $d_2=c_1+c_2$ and you will get the answer from wolfram. 
A: Apart from missing the absolute sign for log. You have the following issues
$$
y = \int \frac{1}{1+t}dt \to \ln(1+t) + C_2
$$
then couple this with the following.
$$
\log(1 + \mathrm{e}^{x+c}) = \log(1 + A\mathrm{e}^{x}) = \log A\left(\frac{1}{A} + \mathrm{e}^{x}\right)  = \log A + \log \left(\frac{1}{A} + \mathrm{e}^{x}\right)
$$
A: Hint.
Calling $z = y'$ we have
$$
z' + z^2-z = 0
$$
Making now $u = \frac 1z\Rightarrow u'+u-1=0\Rightarrow u = 1+C_0 e^{-x}$ then
$$
z = \frac{1}{u} = \frac{e^x}{e^e+C_0}\Rightarrow y = \int z dx = \ln(e^x+C_0) + C_1
$$
A: Note that $$ y=\int \frac{e^{x+c}}{1+e^{x+c}}dx$$
$$t=e^{x+c} \implies dt =e^{x+c}dx$$
$$ y=\int \frac{e^{x+c}}{1+e^{x+c}}dx =\int \frac{dt}{1+t} $$
$$ = \ln(1+t)+c_2 = \ln(1+e^{x+c} ) +c_2$$
Their answer $$ ln(c_1+e^x)+c_2 = \ln c_1(1+e^x/{c_1})+c_2=\ln c_1 + ln (1+c_1^{-1} e^x)+c_2$$ is equivalent to your solution via renaming the constants. 
