MCQ
$\int_{}^{} {\frac{{\log x}}{{{{(1 + \log x)}^2}}}dx = } $
- A$\frac{1}{{1 + \log x}} + c$
- B$\frac{x}{{{{(1 + \log x)}^2}}} + c$
- ✓$\frac{x}{{1 + \log x}} + c$
- D$\frac{1}{{{{(1 + \log x)}^2}}} + c$
Put $1 + \log x = t \Rightarrow \frac{1}{x}dx = dt$
$ \Rightarrow dx = x\,dt = {e^{t - 1}}dt,$ then it reduces to
$\int_{}^{} {\frac{{(t - 1)\,{e^{t - 1}}}}{{{t^2}}}dt} = \int_{}^{} {{e^{t - 1}}\left( {\frac{1}{t} - \frac{1}{{{t^2}}}} \right)\,dt} = \frac{{{e^{t - 1}}}}{t} = \frac{x}{{1 + \log x}} + c$.
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Consider the following two statements:
$(A): f(x) \leq 1$, for all $x \in[2,4]$
$(B)$ : $f(x) \geq \frac{1}{8}$, for all $x \in[2,4]$
Then,