Answer
$$\ln (2t+1)- \ln (t+1)+c $$
Work Step by Step
Given
$$\int \frac{dt}{2t^2+3t+1}$$
Since
\begin{align*}
\frac{1}{2t^2+3t+1}&=\frac{A}{2t +1}+\frac{B}{ t +1}\\
&=\frac{ A(t+1)+ B(2t+1)}{2t^2+3t+1}\\
1&=A(t+1)+ B(2t+1)
\end{align*}
At $t=-1 \ \ \Rightarrow B=-1$, and At $t=-1/2 \ \ \Rightarrow A=2$, then
$$\frac{1}{2t^2+3t+1} =\frac{2}{2t +1}-\frac{1}{ t +1}$$
Hence
\begin{align*}
\int\frac{dt}{2t^2+3t+1}&=\int\frac{2dt}{2t +1}-\int\frac{dt}{ t +1}\\
&=\ln (2t+1)- \ln (t+1)+c
\end{align*}
