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