Discussion
\(\displaystyle\int \frac{\log(1 + x)}{(1 + x)}\, dx =\)
Answer: B
The integral is \(\frac{1}{2}[\log(1 + x)]^{2} + C\).
Substitute \(t = \log(1 + x)\), so \(dt = \frac{dx}{1 + x}\):
\[ \begin{aligned} \int t\,dt &= \frac{t^{2}}{2} + C \\ &= \frac{1}{2}[\log(1 + x)]^{2} + C \end{aligned} \]
Without the \(\frac{1}{2}\) the derivative would be twice the integrand. \(\frac{1}{2}\log(1 + x)\) and \(\log(1 + x)\) are the integrals of \(\frac{1}{1 + x}\)-type functions, not of this one.
The general pattern is \(\int f(x)f'(x)\,dx = \frac{[f(x)]^{2}}{2} + C\), here with \(f(x) = \log(1 + x)\).
No comment is present. Be the first to comment.
Loading…
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