Question

Integrate the function w..r. to x : $\frac{x^{2}+2}{(x-1)(x+2)(x+3)}$

Answer 1

Solution:The given problem can be solved by partial fraction.
$\frac{x^{2}+2}{(x-1)(x+2)(x+3)}= \frac{A}{(x-1)}+\frac{B}{(x+2)}+\frac{C}{(x+3)}$

To find the values of constant

$A= \frac{x^{2}+2}{(x+2)(x+3)}\:\:put x=1\\\\A=\frac{3}{12}\\\\A=\frac{1}{4}\\\\B= \frac{x^{2}+2}{(x-1)(x+3)}\:\:put x=-2\\\\B=\frac{6}{-3}\\\\B=-2\\\\C= \frac{x^{2}+2}{(x+2)(x-1)}\:\:put x=-3\\\\C=\frac{11}{4}$

So
$\frac{x^{2}+2}{(x-1)(x+2)(x+3)}= \frac{1}{4(x-1)}+\frac{-2}{(x+2)}+\frac{11}{4(x+3)}$

Now integrate
$\int\frac{x^{2}+2}{(x-1)(x+2)(x+3)} dx= \int \frac{1}{4(x-1)} dx+\int\frac{-2}{(x+2)} dx+\int\frac{11}{4(x+3)} dx\\\\=\frac{log (x-1)}{4}-2 log (x+2)+\frac{11 ×log( x+3)}{4}+C$