A Super-Dumb question !

[Onixa]

What is the integral of 1/(ax+b) ?
I know I know it is easy and is solved like this:

int[1/(ax+b)]dx
= 1/a int[a/(ax+b)]dx
= 1/a ln(ax+b)

But hehe you know what guys, here is an another solution:

int[1/(ax+b)]dx
= 1/a int[1/x+(b/a)]dx
= 1/a ln[x+(b/a)]

But 1/a ln(ax+b) <> 1/a ln[x+(b/a)]

What the **** is going on ???

 

[HallsofIvy]

It may be easy but neither of those is correct! You have not included the "constant of integration".
Yes, \(\displaystyle \int\frac{dx}{ax+ b}= \frac{1}{a} ln|ax+ b|+ C_1\) where I have included the absolute value (1/(ax+ b) is defined for ax+b< 0 but ln(ax+b) is not) and the constant of integration.

Yes, \(\displaystyle \frac{1}{ax+ b}= \frac{1}{a}\frac{1}{x+ b/a}\). That is correct. And integrating that last, you get \(\displaystyle \frac{1}{a}ln|x+ b/a|+ C_2\) where "\(\displaystyle C_2\)" is the constant of integration, not necessarily the same as \(\displaystyle C_1\).

Now, (1/a)ln|x+ b/a|= (1/a)ln(|1/a||ax+ b|)= (1/a)ln|ax+ b|+ (1/a)ln(1/a)= (1/a)ln|ax+ b|+ C where C= (1/a)ln(1/a) is a constant. That is, your two solutions differ by a constant, which can be included in the "constant of integration".

 

[Onixa]

It may be easy but neither of those is correct! You have not included the "constant of integration".
Yes, \(\displaystyle \int\frac{dx}{ax+ b}= \frac{1}{a} ln|ax+ b|+ C_1\) where I have included the absolute value (1/(ax+ b) is defined for ax+b< 0 but ln(ax+b) is not) and the constant of integration.

Yes, \(\displaystyle \frac{1}{ax+ b}= \frac{1}{a}\frac{1}{x+ b/a}\). That is correct. And integrating that last, you get \(\displaystyle \frac{1}{a}ln|x+ b/a|+ C_2\) where "\(\displaystyle C_2\)" is the constant of integration, not necessarily the same as \(\displaystyle C_1\).

Now, (1/a)ln|x+ b/a|= (1/a)ln(|1/a||ax+ b|)= (1/a)ln|ax+ b|+ (1/a)ln(1/a)= (1/a)ln|ax+ b|+ C where C= (1/a)ln(1/a) is a constant. That is, your two solutions differ by a constant, which can be included in the "constant of integration".

Yes that's true !! The constant of integration does the trick !
Thank you HallsofIvy