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

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
