Question

ʃ x-3/(x+6)^3 dx how do u integrate this ecuation by repeated linear factor

Answer 1

(9-2x)/(2*(x-6)^2)
use integration by parts

Answer 2

I have a better way.\[\frac{x-3+3-3}{(x-6)^3} = (x-6)^2 + \frac 3{(x-6)^3}\]
Now integrate it.

Answer 3

\[\int\limits \frac{x - 3}{(x + 6)^3}dx\]\[u = x + 6\]\[x = u - 6\]\[dx = du\]\[\int\limits \frac{u - 9}{u^{3}}du= \int\limits (u^{-2} - 9u^{-3}) du\]

Answer 4

@Ishaan94 where did the +3-3 come from in the numerator?

Answer 5

curious because i'll be in this section soon

Answer 6

He just add 0 (+3 - 3), because them we can bundle it like this: \[ \frac{x-6 + 3}{(x-6)^3} = \frac{x-6}{(x-6)^3} + \frac{3}{(x-6)^3}\]that yields what he got. Although, I think it should be a -2, rather than a 2.

Answer 7

ah i see thank you for the explanation@bmp

Answer 8

No problem.