Why is if you integrate 1 over (100-v) you get -ln(100-v)

Ste234

Why is if you integrate 1 over (100-v) you get -ln(100-v)
I understand 1/x is ln(X) but why is the above a negative?

dambarude

If you don't get an answer here you should post in the Maths forum. It's been a while since I integrated so I can't answer your question.

Ste234

Oh yeah, Thanks.

**Timbuk2**

[latex]\displaystyle \int \frac{1}{100-v}\,\, dv[/latex]

To integrate this, you use the fact that [latex]\int\frac{1}{x}[/latex] = ln|x| but you also have to remember that you divide by the derivative of x (in this case v). Remember, if you were differentiating, you would multiply by the derivative, so if you're intergrating, you divide.

Therefore
[latex]ln(100-v).(-1)[/latex] as the derivative of -v is -1
[latex]-ln(100-v)[/latex] (leaving it in this form would be fine)
[latex]ln(\frac{1}{100-v})[/latex]

If you're still confused, go by this formula (not in the new tables!!)
[latex]\displaystyle \int\frac{1}{ax+b}\,\, = \,\, \displaystyle \frac{1}{a}ln(ax+b)[/latex]

I hope this helps!

Ste234

That's perfect thank you. That helps a lot I could remember why it does.

Making It Bad

You can use a substitution as well which is how I prefer but it can take longer.

[latex]\displaystyle \int \frac{1}{100-v}\,\, dv[/latex]

Let [latex] {u}={100-v} [/latex]

Therefore[latex] \frac {du}{dv}={-1}[/latex] (differentiating both side with respect to v)

[latex]{du}={-dv} [/latex] (multiply both sides by dv}
Substituting back into [latex]\displaystyle \int \frac{1}{100-v}\,\, dv[/latex]we get [latex]\displaystyle \int \frac{-1}{u}\,\, du[/latex] take out the -1 gives us [latex] -1 \displaystyle \int \frac{1}{u}\,\, du[/latex]

Integrating gives you [latex] {(-1)}{ln (u)} [/latex] and substituting back in for u gives us [latex] {-ln(100-v)} [/latex]

It's the same thing really I guess but sometimes I find it hard to do the way **Timbuk2** does and like to do a substitution when the questions get more difficult.