1. ## Re: Integration help required~~

So when I integrated them, I got
ln (u) -6ln(1+6u)-6ln(1/1+6u)

is this right? if it's right do i just subsitute the u back to the original value?

2. ## Re: Integration help required~~

Let's back up a step, are you sure your partial fraction decomposition is correct?

Once you are sure you have the correct decomposition, when you integrate, make the limits of integration reflect the substitution and you won't need to back-substitute for the original variable.

3. ## Re: Integration help required~~

I checked my A,B and C and C should actually be -6 instead of 6. is that right? And if it's right do i substitute my u with my upper and lower limit?

4. ## Re: Integration help required~~

Yes, C = -6. Then if $x_1$ and $x_2$ are the original lower and upper limits respectively, change the limits to $u(x_1)$ and $u(x_2)$.

5. ## Re: Integration help required~~

My upper limit is 12 and lower limit is so i substituted them as (ln12-6ln(1+6(12))-6ln(1/1+6(12))-(ln7-6ln(1+6ln(7)-6ln(1/1+6(7)) and the answer I got is 0.539 (in 3 sig fig). What did i do wrong? I used the graphics calculator to check the answer of the original function and it is 97.1 ... What did i do wrong? did I jump a step?

6. ## Re: Integration help required~~

Originally Posted by ghfjdksla94
So when I integrated them, I got
ln (u) -6ln(1+6u)-6ln(1/1+6u)

is this right? if it's right do i just subsitute the u back to the original value?
First of all, you need absolute value signs around the stuff inside the logarithms.

The final term is also incorrect. \displaystyle \begin{align*} \frac{1}{(1 + 6u)^2} = (1 + 6u)^{-2} \end{align*}, now integrate...

7. ## Re: Integration help required~~

Then, is the correct term -(1/6)(1+6u)^-1 ? and by absolute values for logarithms, you mean l l <-- these lines around them?

8. ## Re: Integration help required~~

Originally Posted by ghfjdksla94
Then, is the correct term -(1/6)(1+6u)^-1 ? and by absolute values for logarithms, you mean l l <-- these lines around them?
Yes, that is correct

9. ## Re: Integration help required~~

so my final equation just before substituting u into my upper and lower limit is:
(ln lul -6ln l1+6ul +1/6 (1+6u)^-1 ) ? Then multiply the answer by 10.5984 which is the constant we brought out to the in front of the integral?

10. ## Re: Integration help required~~

Originally Posted by ghfjdksla94
so my final equation just before substituting u into my upper and lower limit is:
(ln lul -6ln l1+6ul +1/6 (1+6u)^-1 ) ? Then multiply the answer by 10.5984 which is the constant we brought out to the in front of the integral?
It should actually be \displaystyle \begin{align*} \ln{|u|} - \ln{|1 + 6u|} + \frac{1}{1 + 6u} \end{align*}.

11. ## Re: Integration help required~~

Originally Posted by Prove It
It should actually be \displaystyle \begin{align*} \ln{|u|} - \ln{|1 + 6u|} + \frac{1}{1 + 6u} \end{align*}.
So, I substitute the u into my upper limit 12 and subtracted it by the same equation this time u substituted into my lower limit 7?

I got an answer of 1.8 X 10^-4. I putted all the logarithm numbers as an absolute... what is wrong now??

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