Understanding Integration Mechanics: Why Can't the Constant Be Taken Out?

[bp_psy]

I understand the mechanics of how this happens but i don't really understand why.

\frac{a}{b}\int\frac{1}{x+c}dx\neq\int\frac{a}{b(x+c)}dx

Why can't the constant be taken out?

 

[l'Hôpital]

Why can't the constant be taken out?

What makes you think it can't be?

 

[bp_psy]

What makes you think it can't be?

This:
\frac{a}{b}\int\frac{1}{x+c}dx=\frac{a}{b}ln(x+c)+C

\int\frac{a}{b(x+c)}dx=\frac{a}{b}ln(bx+bc)+C

 

[l'Hôpital]

Consider
<br /> \int 5x<br />

So, which is right?

<br /> \int 5x = \frac{(5x)^2}{2} + C<br />
or
<br /> \int 5x = 5\frac{x^2}{2} + C<br />
Even in regular integration, you always pull off the constants. Just because you have 1/x doesn't mean the constant shouldn't be pulled out.

However, it does worth mentioning that both your answers are actually right.

<br /> \frac{a}{b}ln(bx+bc)+C = \frac{a}{b}ln(b(x+c))+C = \frac{a}{b}ln(x+c)+ \frac{a}{b}ln b + C = \frac{a}{b}ln(x+c)+D<br />
Where D is just another constant.

 

[bp_psy]

Consider
<br /> \int 5x<br />

So, which is right?

<br /> \int 5x = \frac{(5x)^2}{2} + C<br />
or
<br /> \int 5x = 5\frac{x^2}{2} + C<br />
Even in regular integration, you always pull off the constants. Just because you have 1/x doesn't mean the constant shouldn't be pulled out.

I know.

However, it does worth mentioning that both your answers are actually right.

<br /> \frac{a}{b}ln(bx+bc)+C = \frac{a}{b}ln(b(x+c))+C = \frac{a}{b}ln(x+c)+ \frac{a}{b}ln b + C = \frac{a}{b}ln(x+c)+D<br />
Where D is just another constant.

Ok now I understand it. I actually run into this problem while trying to calculate integration factors for ODEs. This should simplify my calculations.

Thank you.