Integral Calculus Question 75
Question: If $ \int{\frac{2x^{2}+3.dx}{(x^{2}-1)(x^{2}-4)}}=\log {{( \frac{x-2}{x+2} )}^{a}}{{( \frac{x+1}{x-1} )}^{b}}+c $ then the values of a and b respectively are
[AMU 2005]
Options:
A) $ \frac{11}{12},\frac{5}{6} $
B) $ \frac{11}{12},\frac{-5}{6} $
C) $ -\frac{11}{12},\frac{5}{6} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ I=\int{\frac{2x^{2}+3}{(x^{2}-1)(x^{2}-4)}} $ $ \frac{2x^{2}+3}{(x^{2}-1)(x^{2}-4)}=\frac{A}{x^{2}-1}+\frac{B}{x^{2}-4} $ $ 2x^{2}+3=A(x^{2}-4)+B(x^{2}-1) $ $ 2x^{2}+3=x^{2}(A+B)-4A-B $ Comparing the coefficient of $ x^{2} $ and constant term on both sides, $ A+B=2 $ …..(i) $ 4A+B=-3 $ …..(ii) On solving both the equations $ A=-\frac{5}{3} $ , $ B=\frac{11}{3} $ $ \int{\frac{2x^{2}+3.dx}{(x^{2}-1)(x^{2}-4)}} $ $ =\int{\frac{-\frac{5}{3}dx}{(x^{2}-1)}+\int{\frac{\frac{11}{3}dx}{(x^{2}-4)}}} $ $ =-\frac{5}{3}\int{\frac{dx}{(x+1)(x-1)}} $ $ +\frac{11}{3}\int{\frac{dx}{(x+2)(x-2)}} $ $ =-\frac{5}{3}.\frac{1}{2}\int{\frac{dx}{x-1}+\frac{5}{6}\int{\frac{dx}{x+1}}}+\frac{11}{3}.\frac{1}{4}\int{\frac{dx}{x-2}-\frac{11}{12}\int{\frac{dx}{x+2}+c}} $ $ =-\frac{5}{6}\log (x-1)+\frac{5}{6} $ log(x+1) $ +\frac{11}{12} $ log(x?2) $ -\frac{11}{12} $ log(x+2)+c $ =\frac{5}{6}\log ( \frac{x+1}{x-1} )+\frac{11}{12}\log ( \frac{x-2}{x+2} )+c $ $ =\log {{( \frac{x+1}{x-1} )}^{5/6}}+\log {{( \frac{x-2}{x+2} )}^{11/12}}+c $
Þ $ a=\frac{11}{12} $ and $ b=\frac{5}{6} $ .
BETA
