Evalute : x^2+x-1 x^2+x-6 d x - Vidyadip

Question

Evalute : $\int \frac{x^2+x-1}{x^2+x-6} d x$

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Answer

$
\text { Let } \begin{aligned}
I & =\int \frac{x^2+x-1}{x^2+x-6} d x \\
& =\int \frac{\left(x^2+x-6\right)+5}{x^2+x-6} d x \\
& =\int\left[1+\frac{5}{x^2+x-6}\right] d x \\
& =\int 1 d x+5 \int \frac{1}{x^2+x-6} d x
\end{aligned}
$
Let $\frac{1}{x^2+x-6}=\frac{1}{(x+3)(x-2)}=\frac{A}{x+3}+\frac{B}{x-2}$
$
\therefore 1= A ( x -2)+ B ( x +3)
$
Put $x+3=0$, i.e. $x=-3$, we get
$
\begin{aligned}
& 1= A (-5)+ B (0) \\
& \therefore A =\frac{-1}{5}
\end{aligned}
$
Put $x-2=0$, i.e. $x=2$, we get
$
\begin{aligned}
& 1= A (0)+ B (5) \\
& \therefore B =\frac{1}{5} \\
& \therefore \frac{1}{x^2+x-6}=\frac{(-1 / 5)}{x+3}+\frac{(1 / 5)}{x-2} \\
& \therefore I=\int 1 d x+5 \int\left[\frac{(-1 / 5)}{x+3}+\frac{(1 / 5)}{x-2}\right] d x \\
& \quad=\int 1 d x-\int \frac{1}{x+3} d x+\int \frac{1}{x-2} d x
\end{aligned}
$

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