We have, $f\left(\frac{t+1}{2 t+1}\right)=t+1$
$$
\begin{aligned}
& \text { Put } \frac{t+1}{2 t+1}=x \Rightarrow t+1=2 t x+x \\
& \Rightarrow \quad 1-x=t(2 x-1) \Rightarrow t=\frac{1-x}{2 x-1} \\
& \therefore \quad f(x)=\frac{1-x}{2 x-1}+1 \Rightarrow f(x)=\frac{1-x+2 x-1}{2 x-1} \\
& \Rightarrow \quad f(x)=\frac{x}{2 x-1} \\
& \therefore \quad \int f(x) d x=\int \frac{x}{2 x-1} d x=\frac{1}{2} \int \frac{2 x}{2 x-1} d x \\
& =\frac{1}{2} \int \frac{2 x-1+1}{2 x-1} d x=\frac{1}{2} \int\left(1+\frac{1}{2 x-1}\right) d x \\
& =\frac{1}{2}\left[x+\frac{1}{2} \log (2 x-1)\right]+c \\
& =\frac{x}{2}+\frac{1}{4} \log (2 x-1)+c \\
&
\end{aligned}
$$