C4 integration

Hi

I am attaching some work I've been doing on integration. I have checked the answer with an on-line integrator and we have a different solutions.

My two bits of working out are at the top of the page and the online solution after the red wiggly line. Why has it decided to rearrange x as it has done? Is there any way of getting my answer the same as the online answer?

Thanks

The online integrator is correct, your answer is not. Even though you've worked out that $dx$ becomes $du/3$, you end up failing to implement this in your transformed integral [which is possibly a symptom of the fact that you forget to write $dx$ in your integrals in the first place, leading to your own confusion].

To be clear, after the substitution you used, the integral should read:

$\displaystyle\int \dfrac{9x}{3x+7} dx = \displaystyle\int \dfrac{3u-21}{u} \dfrac{du}{3} = \displaystyle\int \dfrac{u-7}{u} du$

And NOT, as you've written:

$\displaystyle\int \dfrac{3u-21}{u} du$

On your second question about the online integrator's method, which is a rather unnatural/non-instructive way to write it, note that the idea is essentially the same as the following observation. [In case it isn't clear, this is not a substitution]:

$\displaystyle\int \dfrac{9x}{3x+7} dx = \displaystyle \int \dfrac{9x+21 - 21}{3x+7} dx = \displaystyle\int \dfrac{3(3x+7)}{3x+7} - \dfrac{21}{3x+7} dx$ $= \displaystyle\int 3 - 7 \times \dfrac{3}{3x+7} dx$

The reason this is helpful is because it reduces the integral to logarithms, which you should be able to see by inspection.

Silly me. and the +7 in mine is just incorporated into +C in theirs I suppose.

Yes. The main message to take from this thread is to remember to write $dx, du$ etc in all integrals. That way, you force yourself to deal with it when making substitutions and thus leave yourself far less likely to make errors of this sort.