Integral $\int \frac{\sqrt{x}}{(x+1)^2}dx$ First, i used substitution $x=t^2$ then $dx=2tdt$ so this integral becomes $I=\int \frac{2t^2}{(1+t^2)^2}dt$ then i used partial fraction decomposition the following way:
$$\frac{2t^2}{(1+t^2)^2}= \frac{At+B}{(1+t^2)} + \frac{Ct + D}{(1+t^2)^2} \Rightarrow 2t^2=(At+B)(1+t^2)+Ct+D=At+At^3+B+Bt^2 +Ct+D$$
for this I have that $A=0, B=2, C=0, D=-2$ 
so now I have 
$I=\int \frac{2t^2}{(1+t^2)^2}dt= \int\frac{2}{1+t^2}dt - \int\frac{2}{(1+t^2)^2}dt$ 
Now,
$$ \int\frac{2}{1+t^2}dt = 2\arctan t$$
and 
$$\int\frac{2}{(1+t^2)^2}dt$$
using partial integration we have:
$$u=\frac{1}{(1+t^2)^2} \Rightarrow du= \frac{-4t}{1+t^2}$$
and $$dt=dv \Rightarrow t=v$$  
so now we have:
$$\int\frac{2}{(1+t^2)^2}dt =\frac{t}{(1+t^2)^2} + 4\int\frac{t^2}{1+t^2}dt = \frac{t}{(1+t^2)^2} + 4\int\frac{t^2 + 1 -1}{1+t^2}dt = \frac{t}{(1+t^2)^2} + 4t -4\arctan t$$
so, the final solution should be:
$$I=2\arctan t - \frac{t}{(1+t^2)^2} - 4t +4\arctan t$$
since the original variable was $x$ we have 
$$I= 6\arctan \sqrt{x} - \frac{\sqrt{x}}{(1+x)^2} - 4\sqrt{x} $$
But, the problem is that the solution to this in my workbook is different, it says that solution to this integral is $$I=\arctan \sqrt{x} - \frac{\sqrt{x}}{x+1}$$
I checked my work and I couldn't find any mistakes, so i am wondering which solution is correct?

Notice, you can continue from here without using partial fractions $$\int \frac{2t^2}{(1+t^2)^2}\ dt$$
 $$=\int \frac{2t^2}{t^4+2t^2+1}\ dt$$
$$=\int \frac{2}{t^2+\frac{1}{t^2}+2}\ dt$$
$$=\int \frac{\left(1+\frac{1}{t^2}\right)+\left(1-\frac{1}{t^2}\right)}{t^2+\frac{1}{t^2}+2}\ dt$$
$$=\int \frac{\left(1+\frac{1}{t^2}\right)dt}{t^2+\frac{1}{t^2}+2}+\int \frac{\left(1-\frac{1}{t^2}\right)dt}{t^2+\frac{1}{t^2}+2}$$
$$=\int \frac{d\left(t-\frac{1}{t}\right)}{\left(t-\frac{1}{t}\right)^2+4}+\int \frac{d\left(t+\frac{1}{t}\right)}{\left(t+\frac{1}{t}\right)^2}$$
$$=\frac 12\tan^{-1}\left(\frac{t-\frac 1t}{2}\right)-\frac{1}{\left(t+\frac{1}{t}\right)}+c$$
$$=\frac 12\tan^{-1}\left(\frac{x-1}{2\sqrt x}\right)-\frac{\sqrt x}{x+1}+c$$
let $\sqrt x=\tan\theta\implies \theta=\tan^{-1}(\sqrt x)$, 
$$=\frac 12\tan^{-1}\left(\frac{\tan^2\theta-1}{2\tan\theta}\right)-\frac{\sqrt x}{x+1}+c$$
$$=\frac 12\tan^{-1}\left(\tan\left(\frac{\pi}{2}+2\theta\right)\right)-\frac{\sqrt x}{x+1}+c$$
$$=\frac{\pi}{4}+\theta-\frac{\sqrt x}{x+1}+c$$
$$=\theta-\frac{\sqrt x}{x+1}+C$$
$$=\color{red}{\tan^{-1}(\sqrt x)-\frac{\sqrt x}{x+1}+C}$$
