Integral Calculus Question 80
Question: If $ I_{n}=\int {{(lnx)}^{n}}dx $ , then $ I_{n}+n{I_{n-1}} $ =
Options:
A) $ \frac{{{(lnx)}^{n}}}{x}+C $
B) $ x{{(lnx)}^{n-1}}+C $
C) $ x{{(lnx)}^{n}}+C $
D) none of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ I_{n}=x{{(\ln x)}^{n}}-\int{\frac{x(n){{(\ln x)}^{n-1}}}{x}}dx $ $ ={{(\ln x)}^{n}}-n{I_{(n-1)}} $ or $ I_{n}+n{I_{n-1}}=x{{(\ln x)}^{n}} $
BETA
