Question

If $\int \frac{2x+3}{(x-1)(x^2+1)}dx=lo{g}_e\left\{\right(x-1{)}^{\frac{5}{2}}(x^2+1{)}^a\}-\frac{1}{2}ta{n}^{-1}x+A$, where A is any arbitrary constant, then the value of ‘a’ is

• A. $\frac{5}{4}$
  
• B. $-\frac{5}{3}$
  
• C. $-\frac{5}{6}$
  
  
• D. $-\frac{5}{4}$

Answer 1

The correct option is D $-\frac{5}{4}$

$I=\int \frac{2x+3}{(x-1)(x^2+1)}dx$
$=\int \frac{5dx}{2(x-1)}+\int \frac{-(\frac{5}{2}x+\frac{1}{2})}{x^2+1}dx$,(By partial fraction)
$I=\frac{5}{2}log(x-1)-\frac{5}{2}\int \frac{xdx}{1+x^2}-\frac{1}{2}\int \frac{dx}{1+x^2}$
$I=\frac{5}{2}log(x-1)-\frac{5}{4}log(1+x^2)-\frac{1}{2}ta{n}^{-1}x+AI=log(x-1{)}^{\frac{5}{2}}(1+x^2{)}^{-\frac{5}{4}}-\frac{1}{2}ta{n}^{-1}x+A$
On comparing, $a=-\frac{5}{4}$