Question 3

Consider the following integral:

$$\int_0^1 \frac{(3t - 1) \hspace{3pt}dt}{(t + 1)(t + 5)}.$$

• Determine the expansion of the integrand $${\frac{3t - 1}{(t + 1)(t + 5) }}$$ in terms of partial fractions.
  
  By the partial fraction rules, we need to find constants $A$ and $B$, such that:
  $$\frac{3t - 1}{(t + 1)(t + 5)} = \frac{A}{t + 1} + \frac{B}{t + 5}.$$
  Mutiplying both sides by $(t + 1)(t + 5)$, we get the equation:
  $$3t - 1 = A(t + 5) + B(t + 1).$$
  • Putting $t = - 5$ gives:
    $$- 16 = - 4B, \hspace{6pt} B = 4.$$

  • Putting $t = - 1$ gives:
    $$- 4 = 4A, \hspace{6pt} A = -1.$$

  So we have:
  $$\frac{3t - 1}{(t + 1)(t + 5)} = \frac{-1}{t + 1} + \frac{4}{t + 5}.$$

• Hence evaluate the integral.
  
  The integral now becomes:
  $$\int_0^1 \frac{(3t - 1) \hspace{3pt}dt}{(t + 1)(t + 5)} = \int_0^... ...ac{4}{t + 5} - \frac{1}{t + 1}\right) \,\, dt = [4\ln(t + 5) - \ln(t + 1)]_0^1$$
  $$= \left[\ln\left(\frac{(t + 5)^4}{t + 1}\right)\right]_0^1 = \ln\left(\frac{6^4}{2}\right) - \ln (5^4)$$
  $$= \ln 648 - \ln(625) = \ln\left(\frac{648}{625} \right) = 0.03613905.$$