Question

\({\bf{1 - 40}}\)- Evaluate the integral.

\(\int {\frac{{dt}}{{2{t^2} + 3t + 1}}} \)

Short answer

\(\int {\frac{{dt}}{{2{t^2} + 3t + 1}}} = \ln |2t + 1| - \ln |t + 1| + C\)

Step-by-step solution

Definition

Integrationis a method of adding or summing up the parts to find the whole. It is a reverse process of differentiation.

Simplify integrand

Consider the indefinite integral\(\int {\frac{{dt}}{{2{t^2} + 3t + 1}}} \)

Factorizing it can be written as:

\(\int {\frac{{dt}}{{2{t^2} + 3t + 1}}} = \int {\frac{1}{{(2t + 1)(t + 1)}}} \)

Apply partial fraction

For partial fraction write the integrand as shown below:

\(\begin{aligned}{l}\frac{A}{{2t + 1}} + \frac{B}{{t + 1}} = \frac{1}{{(2t + 1)(t + 1)}}\\A(t + 1) + B(2t + 1) = 1\\t(A + 2B) + (A + B) = 1\\A + 2B = 0,A + B = 1\end{aligned}\)

\( \Rightarrow A = 2,B = - 1\)

Evaluate integral

Substitute value of\(A\)and\(B\)we have:

\(\frac{A}{{2t + 1}} + \frac{B}{{t + 1}} = \frac{2}{{2t + 1}} - \frac{1}{{t + 1}}.\)

Integrating we have:

\(\int {\left( {\frac{2}{{2t + 1}} - \frac{1}{{t + 1}}} \right)} dt = \ln |2t + 1| - \ln |t + 1| + C\)

Therefore, the value of the given integral\(\int {\frac{{dt}}{{2{t^2} + 3t + 1}}} = \ln |2t + 1| - \ln |t + 1| + C\).