Question: Can someone explain me what I get wrong here ?

I have this integral
\begin{equation}
\int \frac{1}{x(2 \ln x+1)} dx
\end{equation}

I set
$ u=\ln x $ and
$ du=\frac{1}{x} dx $
and so I obtain
\begin{equation}
\int \frac{1}{2u+1} du 
\end{equation}
and then the solution seems to be
\begin{equation}
\ln|2u+1|=\ln|2 \ln x+1|+c
\end{equation}
However, the correct solution is
\begin{equation}
\frac{\ln|2 \ln x+1|}{2} +c
\end{equation}

Answer: 
> 
> To find the solutions of $x′′+2x′+x=0$ such that x(1)=0. My approach is using the change $y=x′$.
> 
> 
> 

You can also choose $y=x'+x \implies y'+y=0$ or rewrite the DE as: $$(xe^t)''=0$$
It is much too complicated, as this is a o.d.e. with constant coefficients. Its characteristic equation is $(\lambda+1)^2=0$. As it has a double root, $\lambda=-1$, a basis of the space of solutions is made up of the functions $\mathrm e^{-t}$ and $t\mathrm e^{-t}$. Therefore the general solution is
$$x(t)=\mathrm e^{-t}(\alpha +\beta t),$$
and there remains to determine the relation between $\alpha$ and $\beta$ so that $x(0)=1$.