# Simple antiderivative question...

• Mar 9th 2009, 12:38 PM
fattydq
Simple antiderivative question...
How do you take the antiderivative of something to the -1 power? Specifically I'm working with (12-10x)^-1. Using normal antideriv rules it would just be (12-10x)^0/0, which is irrational, could someone answer this for me? Thanks
• Mar 9th 2009, 12:42 PM
skeeter
Quote:

Originally Posted by fattydq
How do you take the antiderivative of something to the -1 power? Specifically I'm working with (12-10x)^-1. Using normal antideriv rules it would just be (12-10x)^0/0, which is irrational, could someone answer this for me? Thanks

antiderivative involves the ln function.

$
\int \frac{1}{12-10x} \, dx = -\frac{1}{10}\ln|12-10x| + C
$

normally, one learns how to find the derivative of ln functions before tackling antiderivatives.
• Mar 10th 2009, 11:31 AM
fattydq
Quote:

Originally Posted by skeeter
antiderivative involves the ln function.

$
\int \frac{1}{12-10x} \, dx = -\frac{1}{10}\ln|12-10x| + C
$

normally, one learns how to find the derivative of ln functions before tackling antiderivatives.

Yeah well my calc I class was a joke, I passed out with an A and now I'm struggling hard with calc II. thanks for the reply though

Quote:

Originally Posted by skeeter
antiderivative involves the ln function.

$
\int \frac{1}{12-10x} \, dx = -\frac{1}{10}\ln|12-10x| + C
$

normally, one learns how to find the derivative of ln functions before tackling antiderivatives.

Where did the -1/10 come from in your solution?
• Mar 10th 2009, 01:45 PM
HallsofIvy
Quote:

Originally Posted by fattydq
Yeah well my calc I class was a joke, I passed out with an A and now I'm struggling hard with calc II. thanks for the reply though

Where did the -1/10 come from in your solution?

He made the substitution u= 12- 10x so du= -10 dx so dx= -(1/10)du.
$\int \frac{dx}{12- 10x}= \int\frac{-(1/10)du}{u}= -\frac{1}{10}\int\frac{du}{u}$

$= -\frac{1}{10}ln|u|+ C= -\frac{1}{10}ln|12-10x|+ C$