Evaluate the following integrals
Hi
Let's find the partial fraction decomposition of $\displaystyle \frac{x}{x^2+x-6}$.
The discriminant of the denominator is $\displaystyle \Delta = 1^2-4\cdot1\cdot (-6)=25>0$ hence the denominator has to different real roots : $\displaystyle \frac{-1\pm\sqrt{25}}{2}$ that is to say -3 and 2. Hence $\displaystyle x^2+x-6=(x+3)(x-2)$
It gives $\displaystyle
\frac{x}{x^2+x-6} = \frac{x}{(x+3)(x-2)}=\frac{\alpha}{x+3}+\frac{\beta}{x-2}$ which has to be solved for $\displaystyle \alpha$ and $\displaystyle \beta$.
Can you take it from here ?
$\displaystyle \begin{aligned}
\frac{x}{x^{2}+x-6}&=\frac{2x+1-1}{2(x-2)(x+3)} \\
& =\frac{1}{2}\left[ \frac{2x+1}{x^{2}+x-6}-\frac{(x+3)-(x-2)}{5(x-2)(x+3)} \right] \\
& =\frac{1}{2}\left[ \frac{2x+1}{x^{2}+x-6}-\frac{1}{5}\left\{ \frac{1}{x-2}-\frac{1}{x+3} \right\} \right].
\end{aligned}$
Integrate.
P.S.: first integral can be tackled by double integration technique too.