Integral Calculus Question 331
Question: $ \int_{{}}^{{}}{x^{n}\log x\ dx=} $
Options:
A) $ \frac{{x^{n+1}}}{n+1}{ \log x+\frac{1}{n+1} }+c $
B) $ \frac{{x^{n+1}}}{n+1}{ \log x+\frac{2}{n+1} }+c $
C) $ \frac{{x^{n+1}}}{n+1}{ 2\log x-\frac{1}{n+1} }+c $
D) $ \frac{{x^{n+1}}}{n+1}{ \log x-\frac{1}{n+1} }+c $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \int_{{}}^{{}}{x^{n}\log x,dx}=\log x,.,\frac{{x^{n+1}}}{n+1}-\int_{{}}^{{}}{\frac{{x^{n+1}}}{n+1},.,\frac{1}{x},dx} $ $ =\frac{{x^{n+1}}}{n+1}\log x-\frac{{x^{n+1}}}{{{(n+1)}^{2}}}+c=\frac{{x^{n+1}}}{n+1}[ \log x-\frac{1}{n+1} ]+c. $
BETA
