<s>[INST] Evaluation of $\int\frac{2+\sqrt{x}}{\left(x+\sqrt{x}+1\right)^2}dx$ Evaluation of $\displaystyle \int\frac{2+\sqrt{x}}{\left(x+\sqrt{x}+1\right)^2}dx$
$\bf{My\; Try::}$ Let $$\displaystyle I = \int\frac{2+\sqrt{x}}{\left(x+\sqrt{x}+1\right)^2}dx\;,$$ Now Put $x=t^2\;,$ Then $dx = 2tdt$
Then Integral $$\displaystyle I = \int\frac{(2+t)\cdot 2t}{(t^2+t+1)^2}dt = 2\int\frac{(2t+t^2)}{(t^2+t+1)^2}dt$$
Now We can write Integral $$\displaystyle I = 2\int\frac{(2t^{-3}+t^{-2})}{(1+t^{-1}+t^{-2})^2}dt\;,$$ 
Now Put $(1+t^{-1}+t^{-2})=u\;,$ Then $(-t^{-2}-2t^{-3})dt=du\Rightarrow (t^{-2}+2t^{-3})dt=-du$
So Integral $$\displaystyle I = -2\int \frac{1}{u^2}dt = \frac{2}{u}+\mathcal{C} = \frac{2t^2}{t^2+t+1}+\mathcal{C} = \frac{2\sqrt{x}}{x+\sqrt{x}+1}+\mathcal{C}$$
My question is why my answer is different from wolframalpha.com
http://www.wolframalpha.com/input/?i=integration+of+%282%2Bsqrt%28x%29%29%2F%28x%2Bsqrt%28x%29%2B1%29%5E2
Thanks
 [/INST] You have a typo at the very last step.
Note that 
$$\frac{2t^2}{t^2+t+1}=\frac{2\color{red}{x}}{x+\sqrt x+1}$$
and that 
$$\frac{2x}{x+\sqrt x+1}-2=\frac{2x-2(x+\sqrt x+1)}{x+\sqrt x+1}=\frac{-2(\sqrt x+1)}{x+\sqrt x+1}$$
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