Integral Calculus Question 94
Question: If $ \int{\frac{(2x^{2}+1)dx}{(x^{2}-4)(x^{2}-1)}=\log
[ {{( \frac{x+1}{x-1} )}^{a}}{{( \frac{x-2}{x+2} )}^{b}} ]}+C, $ then the values of a and b are respectively [Roorkee 2000]
Options:
A) 1/2, ¾
B) -1, 3/2
C) 1, 3/2
D) -1/2, ¾
Show Answer
Answer:
Correct Answer: A
Solution:
$ I=\int{\frac{(2x^{2}+1)}{(x^{2}-4)(x^{2}-1)}dx} $
$ \frac{2x^{2}+1}{(x^{2}-4)(x^{2}-1)}=\frac{3}{(x^{2}-4)}-\frac{1}{x^{2}-1} $
$ \therefore I=\int{[ \frac{3}{(x^{2}-4)}-\frac{1}{x^{2}-1} ]}dx $
$ =\frac{3}{2\times 2}\log | \frac{x-2}{x+2} |-\frac{1}{2}\log | \frac{x-1}{x+1} |+c $
$ =\frac{3}{4}\log | \frac{x-2}{x+2} |+\log {{| \frac{x+1}{x-1} |}^{1/2}}+c $
$ =\log {{| \frac{x-2}{x+2} |}^{3/4}}+\log {{| \frac{x+1}{x-1} |}^{1/2}}+c $
$ =\log [ {{( \frac{x+1}{x-1} )}^{1/2}}{{( \frac{x-2}{x+2} )}^{3/4}} ]+c $ .
BETA
