Question

If $\int \frac{(2x^2+1)dx}{(x^2-4)(x^2-1)}=log\left[{\left(\frac{x+1}{x-1}\right)}^a{\left(\frac{x-2}{x+2}\right)}^b\right]+C$, then the values of a and b are respectively

• A. $\frac{1}{2},\frac{3}{4}$
  
• B. $-1,\frac{3}{2}$
  
• C. $1,\frac{3}{2}$
  
• D. $-\frac{1}{2},\frac{3}{4}$

Answer 1

The correct option is A $\frac{1}{2},\frac{3}{4}$


$I=\int \frac{(2x^2+1)}{(x^2-4)(x^2-1)}dx\frac{2x^2+1}{(x^2-4)(x^2-1)}=\frac{3}{(x^2-4)}-\frac{1}{x^2-1}$
$\therefore I=\int [\frac{3}{(x^2-4)}-\frac{1}{x^2-1}]dx=\frac{3}{2\times 2}log\left|\frac{x-2}{x+2}\right|-\frac{1}{2}log\left|\frac{x-1}{x+1}\right|+c$
$=\frac{3}{4}log\left|\frac{x-2}{x+2}\right|+log{\left|\frac{x+2}{x-2}\right|}^{\frac{1}{2}}+c$
$=log{\left|\frac{x-2}{x+2}\right|}^{\frac{3}{4}}+log{\left|\frac{x+2}{x-2}\right|}^{\frac{1}{2}}+c$
$=log\left[{\left(\frac{x+1}{x-1}\right)}^{\frac{1}{2}}{\left(\frac{x-2}{x+2}\right)}^{\frac{3}{4}}\right]+c$.