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?
 A: $$I=\int\frac{x^{\frac{1}{2}}}{1+x^2}$$
$x=t^2$, 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.$\int \frac{dx}{x^2-z^2}=\frac{1}{2a}\ln(\frac{x-a}{x+a})$
:)
A: $$I = \int \dfrac{x^{1/2}}{1+x^2} dx$$ Let $x = t^2$. We get $$I = \int \dfrac{2t^2 dt}{1+t^4}$$ Now factorize $(1+t^4)$ as $(t^2 + \sqrt{2}t+1)(t^2 - \sqrt{2}t+1)$ and use partial fractions.
$$I = \dfrac{1}{\sqrt{2}} \int \left( \dfrac{t}{t^2 - \sqrt{2} t+1} - \dfrac{t}{t^2 + \sqrt{2} t+1}\right)$$
Now $$\int \dfrac{t}{(t-a)^2 + b^2} dt = \int \dfrac{t-a+a}{(t-a)^2 + b^2} dt = \int \dfrac{t-a}{(t-a)^2 + b^2} dt + \int \dfrac{a}{(t-a)^2 + b^2} dt \\= \frac12 \log((t-a)^2+b^2) + \frac{a}{b} \arctan \left( \dfrac{t-a}{b}\right)$$
In our case, $a= \pm \dfrac1{\sqrt{2}}$ and $b = \dfrac1{\sqrt{2}}$.
Hence, the integral is $$\frac1{\sqrt{2}} \left( \frac12 \log(t^2 - \sqrt{2}t + 1) + \arctan(\sqrt{2}t+1) - \frac12 \log(t^2 + \sqrt{2}t + 1) + \arctan(\sqrt{2}t-1) \right) + C$$
Now plug in $t = \sqrt{x}$ to get the integral in terms of $x$.
