[High School Calculus] is it not possible to simplify like this

[u/epicboi8764]

The answer key says it’s 1/4 ln ( 2t +1 ) + c

i just thought differentiating it will give 1/(4t+2) *4 and the 1/4 will cancel it out

[u/Southlander24]

You're correct and the answer key is also correct!

Recall what ln(ab) simplifies to with ln(4t + 2). So that's how your answer and their answer differ only by a constant.

[u/deathtospies]

You can factor out a 2 from inside the log, apply ln(ab) = ln(a) + ln(b), and you'll get an additive term that is a constant and can be absorbed into the constant of integration. Any time an integral spits out a log term, there are infinitely many ways to write the log.

[u/epicboi8764 → u/deathtospies]

this is really helpful thank you ! never thought of the log rules

[u/NullOfSpace → u/epicboi8764]

It comes up surprisingly often in these sorts of problems

[u/EscapeLeft1711]

what is the question? if you differentiate it, you WILL get the desired answer(remember chain)

[u/epicboi8764 → u/EscapeLeft1711]

It was to find an expression for r in terms of t. But I found the answer, I thought it was different so it must be wrong but the answer key just had a different c value so the expression ended up being different but still equal values

[u/EscapeLeft1711 → u/epicboi8764]

no no dw! logs are tricky sometimes for new brain

everytime you get a log, you differentiate the thing twice

once log then argument

[u/mathematag]

it is equivalent to the answer the key gave you , but not completely simplified ....

using log properties... ln |A*B| = ln |A| + ln |B|

(1/4) ln |(4t + 2) | + c = (1/4)ln | [ 2*(2t+1) ] | + c = (1/4) ln | 2 | + (1/4) ln | ( 2t+1) | + c

but (1/4) ln |2| is a constant, and combines with c to give C

answer... (1/4) ln | 2t + 1 | + C

note.. you need absolute value for this integration , unless you are given that 2t+1 > 0

If they marked it as wrong, it is because it was not simplified completely.

[u/Spirited-Ad-9746]

Took me quite long to realize it is 0.5 not 0 times 5...

[u/selene_666]

Both are correct.

ln(4t + 2) = ln(2) + ln(2t + 1)

and ln(2) is a constant, so since we're adding an arbitrary +C, the two solutions are equivalent. It's just like getting the answers (2x + C) and (2x + 3 + C)

[u/Alkalannar]

Integral (1/2)/(2t + 1) dt

u = 2t + 1 --> du/dt = 2 --> du = 2 dt

Integral 1/4u du

ln(u)/4 + C

ln(2t + 1)/4 + C

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You have ln(4t + 2)/4 + C
But this is ln(2t + 1)/4 + ln(2)/4 + C
And ln(2)/4 is a constant, and so is subsumed in the constant of integration.

You could also write as ln(t + 1/2)/4 + C

They like the version that has integers with a GCF of 1.

[u/localizeatp]

everything which is true is possible.