c4 maths, integration confusion

Hey! Hopefully the picture will illustrate my confusion, would really appreciate some help here.
I tried the two different methods of integrating 1/5theta and they don't end up with the same answer..?

thanks for the help in advance and sorry for the bad writing

Good spot. Notice that $\ln |5\theta|+c$ can be split into $\ln 5 + \ln |\theta|+c$

Since $\ln 5$ is a constant, it can be combined with the arbitary constant $c$ to end up with $\ln|\theta| + c$.

So both answers are actually the same. Make sense?

OHH thanks, I would never have seen that without your help.

So in these sorts of questions if I didn't separate the ln(5theta) and carried on with said question, I would lose marks?

I think regardless I'm always going to remember to take out the constant before integrating instead of making it harder haha XD

No you wouldn't lose marks since both answers are correct.

If you integrated $\displaystyle \int 2x \ dx$ and wrote e.g. $x^2+e+c$ then you'd still get the marks because it's correct even though there's a random $e$ in your answer!

Of course I wouldn't advise this - keep your answer as simple as possible but $\ln |5\theta|+c$ is fine.

Hey! Hopefully the picture will illustrate my confusion, would really appreciate some help here.
I tried the two different methods of integrating 1/5theta and they don't end up with the same answer..?

thanks for the help in advance and sorry for the bad writing

here's why u r wrong you are try in to use a statement which states that ;
Integral (f'(x)/f (x)) = ln |f (x)|

your first .met his is correct but the second method if wrong since let's say f (theta) = 5theta then f'(theta) = 5 so u can't use the statement since the numerator is not a derivative of 5 theta.......

let me know if u understood this

No his second method is correct (he's just missing a +c).

You're right that $\frac{1}{5\theta}$ is not exactly in the form you mentioned but that's why he adjusted the answer by multiplying by $\frac{1}{5}$.

$\displaystyle \int \frac{1}{5\theta} \ d\theta = \frac{1}{5}\ln(5\theta) + c$

This is correct.

aha I just saw that now but would that be recommended?

Yes it’s fine. You can always simplify the final answer if needed.

I see...Yeah pretty works at higher levels here at high school they don't bother looking for ways to approach a question like in my pure math class when we were learning integration by parts I tried to use the formula in this way;
Integral v*dy/dx = uv - integral u*dv/dx

it lead me to the same answer but my teacher game me a wrong at the start thinking since my values were at wrong place and he is like "you shouldn't confuse things use what u r given". This discourages students from trying new things and shows that most of the time the teachers don't look at your method instead they look at the final answer directly......is it the same way where you come from? like is this a problem worldwide or only a few teachers are like this?

do they take marks off if you forget to put the modulus signs ?

ln|5x|

do they take marks off if you forget to put the modulus signs ?

ln|5x|

No and they also never ask questions like this

$\displaystyle \int^{-2}_{-3} \frac{1}{x} \ dx$

A cruel part of me wishes they did put something like this in the exam one year - then teachers would have to encourage their students to use modulus signs

here's why u r wrong you are try in to use a statement which states that ;
Integral (f'(x)/f (x)) = ln |f (x)|

your first .met his is correct but the second method if wrong since let's say f (theta) = 5theta then f'(theta) = 5 so u can't use the statement since the numerator is not a derivative of 5 theta.......

let me know if u understood this

No and they also never ask questions like this

$\displaystyle \int^{-2}_{-3} \frac{1}{x} \ dx$

A cruel part of me wishes they did put something like this in the exam one year - then teachers would have to encourage their students to use modulus signs

doesn't that give the same answer regardless of the modulus signs? tried it but maybe i'm wrong i got ln(2/3)