1. ## Integration

Alright so I'm a little confused with basic integration.

$\displaystyle \int\frac{dx}{x}=ln|x|+c$
But what if theres a constant infront of dx.
For example
$\displaystyle \int\frac{7dx}{x}=\int\frac{1}{x}*7dx=?$

What changes then?

2. As 7 is a constant, it is not affected by the integration, so you can do this:

$\displaystyle \int\frac{7dx}{x}$

$\displaystyle = 7\int\frac{dx}{x}$

$\displaystyle = 7ln|x| + c$
$\displaystyle = ln |x^7| + c$

or if you like to think of it this way:

$\displaystyle 7\int\frac{7dx}{x}(\frac{1}{7})$

3. Originally Posted by Gusbob
As 7 is a constant, it is not affected by the integration, so you can do this:

$\displaystyle \int\frac{7dx}{x}$

$\displaystyle = 7\int\frac{dx}{x}$

$\displaystyle = 7ln|x| + c$
$\displaystyle = ln |x^7| + c$

or if you like to think of it this way:

$\displaystyle 7\int\frac{7dx}{x}(\frac{1}{7})$
Thanks, another quick question,
$\displaystyle \int\frac{7dx}{4x+2}=?=7ln|4x+2|$
Or do I have to multiply it by the derivative of 4x+2?

4. $\displaystyle \int\frac{7dx}{4x+2}$

You multiply by the derivative, but don't forget to divide it.

For this one, I'll take the 7 out again and use the substitution u = 4x + 2. du = 4 dx.

You need a 4 on top, so multiply by $\displaystyle \frac{4}{4}$.

$\displaystyle 7\int\frac{4dx}{4x+2}(\frac{1}{4})$

Since 1/4 is a constant, I took it out like I did with the 7. (Note: only constants that are multiplied or divided can be taken out, addition or subtraction of constants are integrated.)

$\displaystyle \frac{7}{4} \int\frac{du}{u}$ (since du = 4dx; u = 4x+2)

=$\displaystyle \frac{7}{4} ln|4x+2| +c$

5. Originally Posted by Gusbob
$\displaystyle \int\frac{7dx}{4x+2}$

You multiply by the derivative, but don't forget to divide it.

For this one, I'll take the 7 out again and use the substitution u = 4x + 2. du = 4 dx.

You need a 4 on top, so multiply by $\displaystyle \frac{4}{4}$.

$\displaystyle 7\int\frac{4dx}{4x+2}(\frac{1}{4})$

Since 1/4 is a constant, I took it out like I did with the 7.

$\displaystyle \frac{7}{4} \int\frac{du}{u}$ (since du = 4dx; u = 4x+2)

=$\displaystyle \frac{7}{4} ln|4x+2| +c$
Thanks a ton, things are a lot clearer now!

6. ## I dont know it this helps

but. $\displaystyle \int\frac{1}{ax+b}dx$=$\displaystyle \frac{1}{a}\ln(ax+b)\$ using the same technique by isolating the derivative of the quantity