Major Help with C3 Integration question Please

Integrate: x/x+3 dx, u=x+3

so du/dx = 1 = dx/du

Subbing u into the equation:
x/u

Rearranging u for x you get u-3=x so therefore (u-3)/u du

What would the next step be? Because I tried doing (u-3)*u^-1 to get 1-3u^-1
then integrating that you get u-3lnu which is (x+3)-3ln(x+3)...

The answer at the back says (x-3)ln(x+3). Any help where I went wrong is greatly appreciated.

You're correct.

Integrate: x/x+3 dx, u=x+3

so du/dx = 1 = dx/du

Subbing u into the equation:
x/u

Rearranging u for x you get u-3=x so therefore (u-3)/u du

What would the next step be? Because I tried doing (u-3)*u^-1 to get 1-3u^-1
then integrating that you get u-3lnu which is (x+3)-3ln(x+3)...

The answer at the back says (x-3)ln(x+3). Any help where I went wrong is greatly appreciated.

Thank you so much. Are you sure? This question really did my head in.

Thank you so much. Are you sure? This question really did my head in.

Yes. (well you are missing an integration constant + c)

If the answer at the back of the book is x - 3ln(x+3) + c then this is also correct, the +c absords the +3.

If the answer at the back of the book is (x-3)ln(x+3) + c as you have written then it is incorrect.

Yes, if you differentiate you have $1 - \frac 3 {x+3} = \frac{x+3 - 3}{x+3} = \frac x {x+3}$.

Oh it does say x-3ln(x+3), my apologies. So is the +3 basically +c? because it did the in the back that the constant is omitted.

How did you get the 1?

Well
x + 3 - 3 ln(x+3) + c1 and
x - 3 ln(x+3) + c2

are both valid integrals, you just have c2 = 3 + c1.

How did you get the 1?

Yes, if you differentiate you have $1 - \frac 3 {x+3} = \frac{x+3 - 3}{x+3} = \frac x {x+3}$.

Oh so both forms are acceptable in the exam?

just differentiating x+3.

As to Zacken's point about arbitrary constant, it doesn't matter if the antiderivatives differ by a constant. If we have $(x+3) - 3\ln(x+3) + C_1$ and $x - 3\ln(x+3) + C_2$, we can say $C_2 = 3 + C_1$, which is what Zacken meant by "absorbed".

This actually crops up a lot with logs. For example, $\displaystyle \int \frac 1 {2x} \mathrm dx$. You can extract $\frac 1 2$ and integrate it to $\frac 1 2 \ln(x) + C_1$, but equally you can substitute $u = 2x$ to get $\frac 1 2 \ln(2x) + C_2$. These are equivalent answers, in the second antiderivative, we have $\displaystyle \frac 1 2 \ln(2x) + C_2 = \frac 1 2 \ln(2) + \frac 1 2 \ln(x) + C_2$. So, we just have $C_1 = \frac 1 2 \ln(2) + C_2$. Note that if you differentiate both, you'll get the same answer, as the constant will differentiate to zero anyway.