antidiffing exponential functinos

[near quartz]

in what case does the coefficient need to be taken out from the denominator??

[ebon steeple]

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[frail sandal]

Don't they just differ by a constant, which is then covered by the constant of integration?

[near quartz]

what i meant was e.g. the first question, how come the value inside ln is x and not 2x even though in the second question, if u implement the same method it is correct

[frail sandal]

$$\frac{1}{2}\ln|2x| =\frac{1}{2}[\ln(2) + \ln|x|] = \frac{1}{2}\ln|x| + \frac{1}{2}\ln(2)$$

[bronze relic]

ℝafain

[frail sandal]

Hence the sets $\left{\frac{1}{2}\ln|x| + C: C\in\mathbb R\right}$ and $\left{\frac{1}{2}\ln|2x|+C: C\in\mathbb R\right}$, as subsets of $Map(\mathbb R,\mathbb R)$, are one and the same

[bronze relic]

ℝafain

[near quartz]

OHH ok thanks

i get what u mean with the constant now hahha

so basically the difference is covered by the +c

is that right

[night rose]

@near quartz has given 1 rep to @frail sandal

[near quartz]

ok thanksss

+close