Discussion
\(\displaystyle\int_{3}^{5} \frac{1}{x(1 + x)}\, dx =\)
Answer: A
The integral is \(\log\left(\frac{10}{9}\right)\).
Use partial fractions \(\frac{1}{x(1 + x)} = \frac{1}{x} - \frac{1}{1 + x}\):
\[ \begin{aligned} \Big[\log\frac{x}{1 + x}\Big]_{3}^{5} &= \log\frac{5}{6} - \log\frac{3}{4} \\ &= \log\frac{20}{18} = \log\frac{10}{9} \end{aligned} \]
\(\log 5\) and \(\log 2\) integrate only one partial fraction, and \(\frac{11}{9}\) and \(\frac{13}{9}\) come from evaluating at the wrong limits.
A quick check: the integrand is about \(\frac{1}{16}\) at the middle of [3, 5], so the integral is about \(2 \times \frac{1}{16} \approx 0.12\), close to \(\log\frac{10}{9} \approx 0.105\) and far below \(\log 2\) or \(\log 5\).
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development