Question
$\int \frac{x^2+x-1}{x^2+x-6} d x$
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Answer
$\text { Let } I =\int \frac{x^2+x-1}{x^2+x-6} d x$
$=\int \frac{x^2+x-6+5}{x^2+x-6} d x$
$=\int\left[1+\left(\frac{5}{x^2+x-6}\right)\right] d x$
Let $\frac{5}{x^2+x-6}=\frac{5}{(x+3)(x-2)}$
$=\frac{A}{x+3}+\frac{B}{x-2}$
$\therefore 5= A (x-2)+ B ( x +3)\ldots(i)$
Putting $x=2$ in (i), we get
$5= B (5)$
$\therefore B =1$
Putting $x=-3$ in (i), we get
$5= A (-5)$
$\therefore A =-1$
$\therefore \frac{5}{(x+3)(x-2)}=\frac{-1}{x+3}+\frac{1}{x-2}$
$\therefore I =\int\left[1+\frac{-1}{x+3}+\frac{1}{x-2}\right] d x$
$=\int d x-\int \frac{1}{x+3} d x+\int \frac{1}{x-2} d x$
$= x -\log | x +3|+\log | x -2|+ c$
$\therefore I =x+\log \left|\frac{x-2}{x+3}\right|+ c $
$=\int \frac{x^2+x-6+5}{x^2+x-6} d x$
$=\int\left[1+\left(\frac{5}{x^2+x-6}\right)\right] d x$
Let $\frac{5}{x^2+x-6}=\frac{5}{(x+3)(x-2)}$
$=\frac{A}{x+3}+\frac{B}{x-2}$
$\therefore 5= A (x-2)+ B ( x +3)\ldots(i)$
Putting $x=2$ in (i), we get
$5= B (5)$
$\therefore B =1$
Putting $x=-3$ in (i), we get
$5= A (-5)$
$\therefore A =-1$
$\therefore \frac{5}{(x+3)(x-2)}=\frac{-1}{x+3}+\frac{1}{x-2}$
$\therefore I =\int\left[1+\frac{-1}{x+3}+\frac{1}{x-2}\right] d x$
$=\int d x-\int \frac{1}{x+3} d x+\int \frac{1}{x-2} d x$
$= x -\log | x +3|+\log | x -2|+ c$
$\therefore I =x+\log \left|\frac{x-2}{x+3}\right|+ c $
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