0 like 0 dislike
26 views
Rewrite $\int \frac{x^3}{(x-2)(x+3)} d x$ in terms of an integral with a numerator that has degree less than 2.
| 26 views

0 like 0 dislike
To do this we use long division of polynomials to discover that
$\frac{x^3}{(x-2)(x+3)}=\frac{x^3}{x^2+x-6}=x-1+\frac{7 x-6}{x^2+x-6}=x-1+\frac{7 x-6}{(x-2)(x+3)},$
so
$\int \frac{x^3}{(x-2)(x+3)} d x=\int x-1 d x+\int \frac{7 x-6}{(x-2)(x+3)} d x .$
The first integral is easy, so only the second requires some work.
by Platinum (141,884 points)

0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
1 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike