the integral of $1/(x\sqrt{4x+1})$

Question

I integrated $\frac{1}{x\sqrt{4x + 1}}$ and got I $\ln\left|\frac{\sqrt{4x+1}-1}{2\sqrt{x}}\right| + K$, where K is a constant term. But my book gives the answer as $\ln\left|\frac{\sqrt{4x+1}-1}{\sqrt{4x+1} + 1}\right| + K$. Where did I make a mistake?

Here is how I solved it.

Answer 1

Let $\sqrt{4x+1}=t$.

Hence, $\frac{4}{2t}dx=dt$ and $x=\frac{t^2-1}{4}$.

Thus, $$\int\frac{1}{x\sqrt{4x+1}}dx=2\int\frac{1}{t^2-1}dt=\int\left(\frac{1}{t-1}-\frac{1}{t+1}\right)dt=\ln\left|\frac{t-1}{t+1}\right|+C$$