Integrating Logarithmic Functions: Explained

[rsq_a]

$$\frac{1}{2}\int \frac{dx}{x+1} = \frac{1}{2} \log(x+1)$$

$$\int \frac{dx}{2x+2} = \frac{1}{2} \log(2x+2)$$

Huh??

 

[Mute]

Those are indefinite integrals, so they're only unique up to an additive constant. Notice:

$$\frac{1}{2}\log(2x+2) = \frac{1}{2}\log(x+1) + \log\sqrt{2}$$

hence the two integrals differ only by a constant, and both are correct.

 

[rsq_a]

Those are indefinite integrals, so they're only unique up to an additive constant. Notice:

$$\frac{1}{2}\log(2x+2) = \frac{1}{2}\log(x+1) + \log\sqrt{2}$$

hence the two integrals differ only by a constant, and both are correct.

Thought never crossed my mind. Thanks.