Answer
$\ln|2t+1|-\ln|t+1|+C$
Work Step by Step
Partial fractons method.
$2t^{2}+3t+1=2t^{2}+2t+t+1=2t(t+1)+(t+1)=(t+1)(2t+1)$
$\displaystyle \frac{1}{2t^{2}+3t+1}=\frac{A}{t+1}+\frac{B}{2t+1}$
$\displaystyle \frac{1}{2t^{2}+3t+1}=\frac{A(2t+1)+B(t+1)}{(t+1)(2t+1)}$
$\displaystyle \frac{1}{2t^{2}+3t+1}=\frac{(2A+B)t+(A+B)}{t+1}\Rightarrow\left\{\begin{array}{ll}
A+B=1 & (1)\\
2A+B=0 & (2)
\end{array}\right.$
$(2)-(1)\Rightarrow A=-1,$
$(1)\Rightarrow B=2$
$\displaystyle \int\frac{dt}{2t^{2}+3t+1}=-\int\frac{dt}{t+1}+\int\frac{2dt}{2t+1}$
$=-\displaystyle \ln|t+1|+2\cdot\frac{1}{2}\ln|2t+1|+C$
$=\ln|2t+1|-\ln|t+1|+C$
