Integration $\int\frac{\sqrt{x+4}}x dx$ by partial fraction

Question

Here's what I came up with: $$\int\frac{\sqrt{x+4}}x dx$$ for $ u = \sqrt { x + 4 } $ $$=\int\frac{u}{u^2-4}\;2u\;dx$$ $$=2\int\frac{u^2}{(u-2)(u+2)}\;dx$$ $$=\frac{A}{u-2}\;+\;\frac{B}{u+2}$$ $$\implies u^2 = A(u+2)+B(u-2)$$ $$\implies u = \pm2\;\implies\;A=1\;and\;B=-1$$ $$=\int\frac{1}{u-2}\;-\;\frac{1}{u+2}\;du$$ $$=2\ln\left|\frac{u-2}{u+2}\right|$$ $$=2\ln\left|\frac{\sqrt{x+4}-2}{\sqrt{x+4}+2}\right|+C$$ However, when I check with the book, they have the answer as: $$2\sqrt{x+4}\;+\;2\ln\left|\frac{\sqrt{x+4}-2}{\sqrt{x+4}+2}\right|+C$$ Did I do something wrong or are these two answer practically the same?

Answer 1

As the highest powers of $u$ in the numerator & in the denominator are same $$\frac{u^2}{(u-2)(u+2)}=A+\frac B{u+2}+\frac C{u-2}$$

Clearly, $A=1$ as the coefficients of the highest powers of $u$ in the numerator & in the denominator are also same