/sci/ - Science & Math » Thread #11112672

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Anonymous Sat 02 Nov 19:09:53 2019 No.11112672 View Reply Original Report

Quoted By: >>11112678 >>11112691 >>11112697

If mathematicians feel the need to add a $+C$ at the end of indefinite integrals, then why isn't the anti-derivative of $\frac{1}{x}$ equal to $ln(Bx) + C$ ? Why is it just $ln(x) + C$ ?

Anonymous

Anonymous Sat 02 Nov 2019 19:13:43 No.11112678 Report

Quoted By: >>11112682

>>11112672
ln(Bx) = ln(B)+ln(x) with B = const.

Anonymous

Anonymous Sat 02 Nov 2019 19:15:14 No.11112682 Report

Quoted By:

>>11112678
goddammit, I'm an idiot. Thanks

Anonymous

Anonymous Sat 02 Nov 2019 19:19:17 No.11112691 Report

Quoted By:

>>11112672
Because if f'(x) = g'(x) everywhere on R, mean value theorem tells us that f(x) = g(x) + c for some constant c.
Git gud, noob

Anonymous

Anonymous Sat 02 Nov 2019 19:23:15 No.11112697 Report

Quoted By: >>11112705

>>11112672
Were the fuck are you getting that B from?

Anonymous

Anonymous Sat 02 Nov 2019 19:25:32 No.11112705 Report

Quoted By:

>>11112697
The derivative of $ln(\alpha x)$ equals $\frac{1}{x}$ for any real value of $\alpha$

Anonymous

Anonymous Sat 02 Nov 2019 19:37:56 No.11112735 Report

Quoted By:

Differentiate ln(ax) wrt x. Don’t forget your chain rule.

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