Integral Calculus Question 484
Question: If $ I_{n}=\int{{{(\log x)}^{n}}dx}, $ then $ I_{n}+n{I_{n-1}}= $
[Karnataka CET 2003]
Options:
A) $ x{{(\log x)}^{n}} $
B) $ {{(x\log x)}^{n}} $
C) $ {{(\log x)}^{n-1}} $
D) $ n{{(\log x)}^{n}} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ I_{n}=\int{{{(\log x)}^{n}}dx} $ …..(i)
$ \therefore {I_{n-1}}=\int{{{(\log x)}^{n-1}}dx} $ …..(ii)
Now, $ I_{n}=\int{{{(\log x)}^{n}}.,dx}={{(\log x)}^{n}}x-n\int{{{(\log x)}^{n-1}}\frac{1}{x}x,dx} $
$ =x{{(\log x)}^{n}}-n\int{{{(\log x)}^{n-1}}dx} $
$ I_{n}=x{{(\log x)}^{n}}-n{I_{n-1}} $ ;
$ \therefore I_{n}+n,{I_{n-1}}=x{{(\log x)}^{n}} $ .
BETA
