Integration

[InterserveVB]

How would I go about this?

Integral 1 / x + 6 dx ?

When I move the x to the top as x^-1 the power becomes zero when I take the antiderivative. That can't be right, can it?

 

[Daniel_Feldman]

Is this 1/(x+6) dx or ((1/x)+6)dx? Clarify and I'll show you how it works.

 

[InterserveVB]

1 / (x+6) sorry my bad

 

[pka]

The derivative of ln(x+6) is 1/(x+6)
So what is the anti-derivative of [1/(x+6)]?

 

[InterserveVB]

ln x i think

 

[Daniel_Feldman]

Substitution.


Let u=x+6

so du=dx


So 1/(x+6) dx=1/u du.

Now, do what pka said.

 

[InterserveVB]

ok so u = x +6 and du = 1 so dx = du

so i bring the u to the top as a u^-1 then when i do the anti derivative I have the same prob it becomes 0 when I add 1

Integration of u^-1 du

 

[Daniel_Feldman]

Again, if the derivative of ln u is 1/u, what is the antiderivative of 1/u du?

 

[InterserveVB]

So, is the answer ln (x+6)?

 

[Daniel_Feldman]

Actually, it's ln(abs(x+6))+C, where abs=absolute value. x+6 can't be negative, because you can't take the natural log of a negative number. Also don't forget the constant of integration. You need it, because this is indefinite integration.


Nice work.

 

[InterserveVB]

Ah. Thx for all your help!