Why do I get different solutions when integrating both sides of an identity?

Question

I am trying to solve an integral, and I get different solutions, but I basically integrate the same funtion.

My problem is based on the following identity, of which I am confident it is correct: $$\frac{1}{2 \sqrt{8} \ \left(x-\sqrt{8}\right)} = \frac{1}{2 \sqrt{8} \ x -16}$$

Now, when I integrate the left side, I get: $$\int \frac{1}{2 \sqrt{8} \ \left(x-\sqrt{8}\right)} \mathrm{d}x = \frac{\sqrt{2}}{8} \ln\left(x-\sqrt{8}\right) + C$$

And when I integrate the right side, I get: $$ \int \frac{1}{2 \sqrt{8} \ x -16} \mathrm{d}x = \frac{\sqrt{2}}{8} \ln\left(\sqrt{32} \ x - 16\right) + C$$

Which is a different solution. What am i doing wrong?

Answer 1

Note that $$ \ln(\sqrt{32} x - 16) = \ln\left(\sqrt{32} (x - \sqrt{8})\right) = \ln(\sqrt{32}) + \ln(x - \sqrt{8}). $$ Hence, the constant term $\frac{\sqrt{2}}{8}\ln(\sqrt{32})$ can be absorbed in the integration constant.

Answer 2

There's nothing wrong. What you got is that $\dfrac{\sqrt{2}}{8} \ln\left(x-\sqrt{8}\right)$ is a primitive of your function, whereas $\dfrac{\sqrt{2}}{8} \ln\left(\sqrt{32} \ x - 16\right)$ is another primitive. Since they differ by a constant, there is no problem.