Question:
The value of \(\int \dfrac{1}{x + x\log x}\,dx\) is
\(\int \dfrac{1}{x + x\log x}\,dx\) का मान क्या है?
Options:
- log(1 + log x)
- x + log x
- 1 + log x
- x log(1 + log x)
- log(1 + log x)
- x + log x
- 1 + log x
- x log(1 + log x)
Correct Answer: a)
Explanation:
Let \(u = 1 + \log x \implies du = \dfrac{1}{x}\,dx\).
\(\int \dfrac{1}{x(1+\log x)}\,dx = \int \dfrac{1}{u}\,du = \log|u| + C = \log(1 + \log x) + C\).
माना \(u = 1 + \log x \implies du = \dfrac{1}{x}\,dx\)।
\(\int \dfrac{1}{u}\,du = \log|u| + C = \log(1 + \log x) + C\)।