Differentiating & Integrating Natural Logs

Hi guys I've gotten myself confused again.

If I integrate

\dfrac{1}{2x+1} dx

I get

1/2ln(2x+1)

Because

\dfrac{1}{2x+1}

Is half of

\dfrac{2}{2x+1}

Which integrates to

ln(2x+1)

But if I'm differentiating a natural log it follows this rule

\dfrac{d}{dx}ln(f(x)) = \dfrac{f'(x)}{f(x)}

So working backwards

\dfrac{d}{dx}(1/2 ln(2x+1))

= 1/2 \dfrac{d}{dx}(ln(2x+1))

= \dfrac{1}{2x+1}

But we're told that:

a*log(x) = log(x^a)

So why isn't

\dfrac{d}{dx}(1/2ln(2x+1)) = \dfrac{d}{dx}(ln\sqrt{2x+1})

I'm really confused

Reply 1

Notnek

Original post by Retsek

But we're told that:

a*log(x) = log(x^a)

So why isn't

\dfrac{d}{dx}(1/2ln(2x+1)) = \dfrac{d}{dx}(ln\sqrt{2x+1})

I'm really confused

Why do you think it's not?

Have you tried differentiating

\ln\sqrt{2x+1}

? What did you get?

Reply 2

RDKGames

Study Forum Helper

Original post by Retsek

So why isn't

\dfrac{d}{dx}(1/2ln(2x+1)) = \dfrac{d}{dx}(ln\sqrt{2x+1})

I'm really confused

It is...?

Just use the u-sub

u=\sqrt{2x+1}

to differentiate

\ln \sqrt{2x+1}

. This gives

\dfrac{1}{u} \cdot (2x+1)^{-1/2} = \dfrac{1}{2x+1}

- but the alternative is clearly neater.

Reply 3

Retsek

Original post by Notnek

Why do you think it's not?

Have you tried differentiating

\ln\sqrt{2x+1}

? What did you get?

Original post by RDKGames

It is...?

Just use the u-sub

u=\sqrt{2x+1}

to differentiate

\ln \sqrt{2x+1}

. This gives

\dfrac{1}{u} \cdot (2x+1)^{-1/2} = \dfrac{1}{2x+1}

- but the alternative is clearly neater.

Okay I've found the problem, I'm as dumb as a box of rocks

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Differentiating & Integrating Natural Logs

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