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202310191053 Exercise 7.9.5

Evaluate the integral

\displaystyle{\int \frac{\mathrm{d}t}{2t^2+3t+1}}.

Extracted from James Stewart. (2012). Calculus (8e)

Roughwork.

\begin{aligned} &\quad\enspace \int\frac{1}{2t^2+3t+1}\,\mathrm{d}t \\ & = \int\frac{1}{(2t+1)(t+1)}\,\mathrm{d}t \\ & = \int\bigg\{ \frac{A}{2t+1} + \frac{B}{t+1}\frac{}{}\bigg\}\,\mathrm{d}t \\ & = \int\frac{A(t+1)+B(2t+1)}{(2t+1)(t+1)}\,\mathrm{d}t\\ & = \int\frac{(A+2B)t+(A+B)}{(2t+1)(t+1)}\,\mathrm{d}t\\ & = \int\bigg\{\frac{2}{2t+1}-\frac{1}{t+1}\bigg\}\,\mathrm{d}t \\ & = \int\frac{2}{2t+1}\,\mathrm{d}t - \int\frac{1}{t+1}\,\mathrm{d}t \\ & = \int\frac{1}{2t+1}\,\mathrm{d}(2t+1) - \int\frac{1}{t+1}\,\mathrm{d}(t+1) \\ & = \ln |2t+1| - \ln |t+1| +C\quad\textrm{for some constant}\\ &= \ln \bigg| \frac{2t+1}{t+1} \bigg| +C \\ \end{aligned}

This problem is not to be attempted.