Solutions - 0060-01

(1) Write the partial fractions general form equation:

$$\frac{1}{(x+2)(x-3)}=\frac{A}{x+2}+\frac{B}{x-3}$$

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(2) Solve for constants.

Cross multiply:

$$\gg \gg 1=A(x-3)+B(x+2)$$

Plug in $x=3$, obtain $1=B\cdot 5$ so $B=1/5$.

Plug in $x=-2$, obtain $1=A(-5)$ so $A=-1/5$.

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(3) Integrate each term:

$$\begin{array}{cc}\int \frac{1}{(x+2)(x-3)}dx& \gg \gg \int \frac{-1/5}{x+2}dx+\int \frac{1/5}{x-3}dx\\ & \\ & \gg \gg -\frac{1}{5}\ln |x+2|+\frac{1}{5}\ln |x-3|+C\end{array}$$