# Math Help - Simple Cal 2 integral

1. ## Simple Cal 2 integral

integral of $\int \, 2x/x-4
$

got a test tomorrow and I have a feeling this may be on the test.

2. Originally Posted by p00ndawg
integral of $2x/x-4
$

got a test tomorrow and I have a feeling this may be on the test.

rewrite as follows or use long division

$2\int \frac{x}{x-4}dx=2\int \frac{(x-4)+4}{x-4}dx=$

$2\int dx + 2\int \frac{4}{x-4}dx=2x+8 \ln|x-4|+c$

3. Originally Posted by TheEmptySet
rewrite as follows or use long division

$2\int \frac{x}{x-4}dx=2\int \frac{(x-4)+4}{x-4}dx=$

$2\int dx + 2\int \frac{4}{x-4}=2x+8 \ln(x-4)+c$
Brilliant.

4. This is in essence exactly the same procedure that the others have used, but with slightly different symbols (my old calc teacher used to call it "tricky" substitution)...

choose u=x-4. This means that du=dx, but it also means that u+4=x. As a result we have the following integral:

$\int \frac{2(u+4)}{u} du$

$
=\int \frac{2u}{u} +\frac{8}{u} du$

$
=2\int du + 8\int \frac{1}{u} du$

$=2u+8\ln{u}+c$

$=2(x-4)+8\ln{|x-4|}+c$

$=2x+8\ln{|x-4|}+C$

5. Originally Posted by teuthid
This is in essence exactly the same procedure that the others have used, but with slightly different symbols (my old calc teacher used to call it "tricky" substitution)...

choose u=x-4. This means that du=dx, but it also means that u+4=x. As a result we have the following integral:

$\int \frac{2(u+4)}{u} du$

$
=\int \frac{2u}{u} +\frac{8}{u} du$

$
=2\int du + 8\int \frac{1}{u} du$

$=2u+8\ln{u}+c$

$=2(x+4)+8\ln{(x-4)}+c$

$=2x+8\ln{(x-4)}+C$

what happens to the 8 when you distribute 2 to x + 4 and why is it x + 4, shouldnt it be x - 4?

6. TheEmptySet & teuthid are missing something important here: it's actually $\ln|x-4|.$

7. your right, it should be x-4. sorry about the typo. The 8 got dumped into the arbitrary constant that is added to the end. Any time you have an antiderivative with spare constants lying around you can do that. For example if your antiderivative was $F(x)=x^2+1+\sin(x)+2+e^x+5$, then the indefinite integral could be written as $x^2+\sin(x)+e^x+C$. You'll notice that I switched from c to C to show this "dumping".