C4 Integration

Hi. Just doing Maths and come across this question with this mark scheme (OCR MEI C4 Jan 06)
I am wondering how comes it is 1/2lny-1/2ln(2-y) instead of 1/2lny+1/2ln(2-y)?

Integrate 1/[2(2-y)]. There's a minus in there.

Differentiating 1/2 * ln (2 - y) creates a minus sign at the front.

Sorry, still don't get it By integrating 1/[2(2-y)], shouldn't it be 1/2*integration of 1/(2-y), which means 1/2ln(2-y) without a minus? :/
Thanks for the reply btw

Sorry, still don't get it By integrating 1/[2(2-y)], shouldn't it be 1/2*integration of 1/(2-y), which means 1/2ln(2-y) without a minus? :/
Thanks for the reply btw

Sorry, still don't get it By integrating 1/[2(2-y)], shouldn't it be 1/2*integration of 1/(2-y), which means 1/2ln(2-y) without a minus? :/
Thanks for the reply btw

x

This doesn't look much like Physics.

We know that $\int \frac{f'(x)}{f(x)} \, \mathrm{d}x = \ln f(x) + c$, yeah? That is, if the numerator is the derivative of the denominator, then integrating it produces the logarithm of the denominator.

In this case, we have: $\displaystyle \frac{1}{2} \int \frac{1}{2-y} \, \mathrm{d}y$ - the numerator is almost the derivative of the denominator - the dervative of the denominator is $\frac{\mathrm{d}}{\mathrm{d}y}(2-y) = -1.$

How do we fix this? Well, $\displaystyle \frac{1}{2} \int \frac{1}{2-y} \, \mathrm{d}y = -\frac{1}{2} \int \frac{-1}{2-y} \, \mathrm{d}y = -\frac{1}{2} \ln |2-y| + c$

since we now have the numerator as the derivative of the numerator.

Try this out on: $\displaystyle \int \frac{x}{2-x^2} \, \mathrm{d}x$ - the numerator is almost the derivative of the numerator.

If you can't spot what to do:

Ahah alright that makes so much sense. I completely forgot about the f'(x)/f(x) thing. Thank you!

No problem! Do you want to test your understanding on the problem I gave at the end?

Yeah is that -1/2ln(2-x^2) if I am not wrong? :/

Yeah is that -1/2ln(2-x^2) if I am not wrong? :/