Integral Calculus Question 148
Question: If $ l^{r}(x) $ means log log log ??.x, the log being repeated r times. then $ \int{{xl(x)l^{2}(x)l^{3}(x)….l^{r}(x)}{{-}^{1}}dx} $ is equal to
Options:
A) $ {l^{r+1}}(x)+C $
B) $ \frac{{l^{r+1}}(x)}{r+1}+C $
C) $ l^{r}(x)+C $
D) None
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Putting $ I_1={l^{r+1}}(x)=t $ and $ \frac{1}{xl(x)l^{2}(x)….l^{r}(x)}dx=dt $ we get, $ \int{\frac{1}{xl^{2}(x)l^{3}(x)…..l^{r}(x)}} $ $ =\int{1.dt=t+C={l^{r+1}}(x)+C} $
BETA
