Thread: A primitive function for simple equation

1. A primitive function for simple equation

Hi!

This shouldnt be hard but for some reason itīs giving me problem.

f(x) = 1 / (x+1)

How do I find the F(x) for this function?
Thanks

2. Hello,

Note that the derivative for (x+1) is 1.

So you have something like u'(x)/u(x), with u(x)=1+x.

The antiderivative of u'(x)/u(x) is ln(|u(x)|)

3. hi

so basically I get ln ( x +1 ) ?

thanks

4. Originally Posted by Twig
so basically I get ln ( x +1 ) ?

thanks

I wonder why you posted this in the Algebra forum? The only way to solve for such a question is to use Calculus techniques.

5. hi

thanks a lot guys

So if I have f(x) = 1 / (2x + 2)
I get F(x) = ln ( 2x + 2 ) ?

Yeah, I should have posted in a different forum, sorry bout that.

6. Not really, because the derivative for 2x+2 is 2, not 1 :-)

Or else you can see that f(x)=(1/2)*1/(x+1)

7. Originally Posted by Twig
thanks a lot guys

So if I have f(x) = 1 / (2x + 2)
I get F(x) = ln ( 2x + 2 ) ?

Yeah, I should have posted in a different forum, sorry bout that.
Because of the Chain Rule, you must carry out the derivative of what's inside the natural log. In this case, it is 2, so $\displaystyle f(x)=\frac{2}{2x+2}$. To make this adjustment for the integral of $\displaystyle f(x)=\frac{1}{2x+2}$, you must divide the integral by 2, so you have an answer of $\displaystyle F(x)=\frac{\ln(2x+2)}{2}$.

8. Originally Posted by Twig
thanks a lot guys

So if I have f(x) = 1 / (2x + 2)
I get F(x) = ln ( 2x + 2 ) ?

Yeah, I should have posted in a different forum, sorry bout that.
I love doing these general cases $\displaystyle \int\frac{a}{bx+c}dx=\frac{a}{b}\int\frac{b}{bx+c} dx$ now that we have the derivative of the quantity we can just integrate integrate integrate! $\displaystyle \frac{a}{b}\ln|bx+c|+C$