<s>[INST] Evaluating $\int\frac{x^{1/2}}{1+x^2}\,dx.$ 
Compute $$\int\frac{x^{1/2}}{1+x^2}\,dx.$$

All I can think of is some integration by substitution. But ran into something scary. Anyone have any tricks?
 [/INST] $$I=\int\frac{x^{\frac{1}{2}}}{1+x^2}$$
$x=t^2$, we get,
$$=\int\frac{2t^2dt}{1+t^4}$$
$$=\int\frac{t^2+1}{1+t^4}dt + \int\frac{t^2-1}{1+t^4}dt$$
upon dividing by $t^2$, we get
$$=\int\frac{1+\frac{1}{t^2}}{(t-\frac{1}{t})^2+2}dt +\int\frac{1-\frac{1}{t^2}}{(t+\frac{1}{t})^2-2}dt$$
All set now, lets integrate
$$\frac{1}{\sqrt2}\tan^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt2}\right)+\frac{1}{2\sqrt2}\ln\left(\frac{t+\frac{1}{t}-\sqrt2}{t+\frac{1}{t}+\sqrt2}\right) +C$$
just replace $t$ with $x^2$ to get the final answer

This yeilds exactly the same answer as Marvis got, (except for the $\tan^{-1}$ part, which  I cannot understand why?)
Exactly the same thing (the constant is also there)

Results used:
1.$\int\frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}(\frac{x}{a})$
2.$\int \frac{dx}{x^2-z^2}=\frac{1}{2a}\ln(\frac{x-a}{x+a})$
:)
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