# indefinite integrals

• June 30th 2006, 12:04 PM
macca101
indefinite integrals
I'm working through my exercise book quite happily finding indefinite integrals using the table of indefinite integrals, I have come across this one and have been staring at it for an hour unable to understand how this answer is derived.

$
\int \frac{-6}{x^4} dx
$

$
\frac{2}{x^3} +c
$
• June 30th 2006, 12:11 PM
CaptainBlack
Quote:

Originally Posted by macca101
I'm working through my exercise book quite happily finding indefinite integrals using the table of indefinite integrals, I have come across this one and have been staring at it for an hour unable to understand how this answer is derived.

$
\int \frac{-6}{x^4} dx
$

$
\frac{2}{x^3} +c
$

Bog standard power integral:

$
\int \frac{-6}{x^4} dx=-6 \int x^{-4} dx=(-6)(-1/3)x^{-3}+c= \frac{2}{x^3} +c
$

RonL
• June 30th 2006, 12:46 PM
macca101
Thanks