1. ## Calc help intergration

can someone tell me how the 2+8 got there and the 4 turned into a du??

$\displaystyle 2\int 1+ \frac{4}{u^2 - 4} du= 2+8 \int \frac{du}{(u+2)(u-2)}$

2. Originally Posted by Legendsn3verdie
can someone tell me how the 2+8 got there and the 4 turned into a du??

$\displaystyle 2\int 1+ \frac{4}{u^2 - 4} du= 2+8 \int \frac{du}{(u+2)(u-2)}$
$\displaystyle 2\int \left( 1+ \frac{4}{u^2 - 4} \right) du=2\int 1du+2\int \frac{4}{u^2-4}du=$

$\displaystyle 2\int du +8\int\frac{1}{u^2-4}du=2\int du +8\int\frac{du}{(u-2)(u+2)}$

3. Originally Posted by TheEmptySet
$\displaystyle 2\int \left( 1+ \frac{4}{u^2 - 4} \right) du=2\int 1du+2\int \frac{4}{u^2-4}du=$

$\displaystyle 2\int du +8\int\frac{1}{u^2-4}du=2\int du +8\int\frac{du}{(u-2)(u+2)}$

hm can u explain with words what is going on i m still a bit not sure what happened there.

4. Originally Posted by Legendsn3verdie
can someone tell me how the 2+8 got there and the 4 turned into a du??

$\displaystyle 2\int 1+ \frac{4}{u^2 - 4} du= 2+8 \int \frac{du}{(u+2)(u-2)}$

could it just because of this.. tell me what you guys think..

$\displaystyle 2(1+4)\int \frac{du}{u^2-4}$

^if this is valid.. that means i pulled a (1+4) across the intergral sign.. is that possible?

5. Originally Posted by Legendsn3verdie
can someone tell me how the 2+8 got there and the 4 turned into a du??

$\displaystyle 2\int 1+ \frac{4}{u^2 - 4} du= 2+8 \int \frac{du}{(u+2)(u-2)}$
this formula is not correct it is missing an integral sign after the 2.

6. Originally Posted by TheEmptySet
this formula is not correct it is missing an integral sign after the 2.
very strange that is wrong.. i got it directly from the solution manual

well here is the orignal problem...

$\displaystyle \int \frac{x^{1/2}}{x-4}dx$

7. Originally Posted by Legendsn3verdie
well here is the orignal problem...

$\displaystyle \int \frac{x^{1/2}}{x-4}dx$
When you did your u sub $\displaystyle u=x^{1/2}$

$\displaystyle \int \frac{x^{1/2}}{x-4}dx=2\int \frac{u^2}{u^2-4}du$

$\displaystyle 2\int \frac{u^2}{u^2-4}du=2\left( \int 1du +\int \frac{4}{u^2-4}du\right)$

Now we distribute the 2 to get

$\displaystyle 2\int du + 2\int \frac{4}{u^2-4}du$

pulling the four out in front of the 2nd integral gives

$\displaystyle 2\int du + (2)4\int \frac{1}{u^2-4}du$

factoring the denominator of the 2nd gives

$\displaystyle 2 \int du + 8 \int \frac{1}{(u-2)(u+2)}du$

Note that these are the same thing

$\displaystyle 8 \int \frac{1}{(u-2)(u+2)}du=8 \int \frac{du}{(u-2)(u+2)}$

I hope this clears it up.

Good luck.

8. Originally Posted by TheEmptySet
When you did your u sub $\displaystyle u=x^{1/2}$

$\displaystyle \int \frac{x^{1/2}}{x-4}dx=2\int \frac{u^2}{u^2-4}du$

$\displaystyle 2\int \frac{u^2}{u^2-4}du=2\left( \int 1du +\int \frac{4}{u^2-4}du\right)$

Now we distribute the 2 to get

$\displaystyle 2\int du + 2\int \frac{4}{u^2-4}du$

pulling the four out in front of the 2nd integral gives

$\displaystyle 2\int du + (2)4\int \frac{1}{u^2-4}du$

factoring the denominator of the 2nd gives

$\displaystyle 2 \int du + 8 \int \frac{1}{(u-2)(u+2)}du$

Note that these are the same thing

$\displaystyle 8 \int \frac{1}{(u-2)(u+2)}du=8 \int \frac{du}{(u-2)(u+2)}$

I hope this clears it up.

Good luck.
ty sir wish i had more reputation power to give you.