Definite Integration Question 278
Question: The value of $ \int_2^{3}{\frac{x+1}{x^{2}(x-1)}dx} $ is
[MP PET 2004]
Options:
A) $ 2\log 2-\frac{1}{6} $
B) $ \log \frac{16}{9}-\frac{1}{6} $
C) $ \log \frac{4}{3}-\frac{1}{6} $
D) $ \log \frac{16}{9}+\frac{1}{6} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ I=\int_2^{3}{\frac{x+1}{x^{2}(x-1)}}dx=\int_2^{3}{( \frac{A}{x^{2}}+\frac{B}{x}+\frac{C}{x-1} )}dx $
$ A(x-1)+B(x)(x-1)+C(x^{2})=x+1 $
Put $ x=0,1,-1, $ we get $ A=-1,B=-2,C=2 $
therefore $ I=-\int_2^{3}{\frac{dx}{x^{2}}-2\int_2^{3}{\frac{dx}{x}+2\int_2^{3}{\frac{dx}{x-1}}}} $
therefore $ I=[ \frac{1}{x} ]_2^{3}-2[\log x]_2^{3}+2[\log (x-1)]_2^{3} $
therefore $ I=\frac{1}{3}-\frac{1}{2}-2\log \frac{3}{2}+2\log 2 $
therefore $ I=\log \frac{16}{9}-\frac{1}{6} $ .
BETA
