Integral Calculus Question 78
Question: If $ \int_{{}}^{{}}{\frac{2x+3}{(x-1)(x^{2}+1)}\ dx={\log_{e}}{ {{(x-1)}^{\frac{5}{2}}}{{(x^{2}+1)}^{a}} }}-\frac{1}{2}{{\tan }^{-1}}x+A $ , where A is any arbitrary constant, then the value of ?a? is
[MP PET 1998]
Options:
A) 5/4
B) - 5/3
C) - 5/6
D) - 5/4
Show Answer
Answer:
Correct Answer: D
Solution:
$ I=\int_{{}}^{{}}{\frac{2x+3}{(x-1)(x^{2}+1)},dx} $
$ =\int_{{}}^{{}}{\frac{5,dx}{2,(x-1)}}+\int_{{}}^{{}}{\frac{-( \frac{5}{2}x+\frac{1}{2} )}{x^{2}+1}},dx $ , (By partial fraction)
$ I=\frac{5}{2}\log (x-1)-\frac{5}{2}\int_{{}}^{{}}{\frac{x,dx}{1+x^{2}}-\frac{1}{2}\int_{{}}^{{}}{\frac{dx}{1+x^{2}}}} $
$ I=\frac{5}{2}\log (x-1)-\frac{5}{4}\log (1+x^{2})-\frac{1}{2}{{\tan }^{-1}}x+A $
$ I=\log {{(x-1)}^{5/2}}{{(1+x^{2})}^{-5/4}}-\frac{1}{2}{{\tan }^{-1}}x+A $
On comparing, $ a=-\frac{5}{4}. $
BETA
