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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?

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Just a general note: In searching for a substitution $x=f(y)$ for $\int \frac{P(x)}{Q(x)}\mathbb{d}x$, one could try to solve $f'(y)=Q(f(y))$ which then leaves one with the formally simpler $\int P(f(y))\mathbb{d}y$. Here, sadly, $P(x)$ is no polynomial and so the resulting integral $\int\tan{(y)}^{1/2}\mathbb{d}y$ is merely shorter but probably not simpler to solve. – NikolajK Jan 23 '13 at 15:04

$$I = \int \dfrac{x^{1/2}}{1+x^2} dx$$ 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 $x$.

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Of course, some people would consider thhis scary. – Gerry Myerson May 23 '12 at 7:15
I think you made a mistake in the second last step, arctan()'s cancel. That $\frac{a}{b}$ part in front of both the arctan()'s- one of it is $\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}=1$, and the other one is $\frac{\frac{-1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}=-1$. – Ashish Gaurav Feb 23 '13 at 7:05
@AshishGaurav Nope. What I have is correct. See Wolframalpha. – user17762 Feb 23 '13 at 19:07
@Jarvis: I got it- You did not combine the final expressions. I re-differentiated it and got the same function. Thanks. – Ashish Gaurav Feb 24 '13 at 4:31


$x=t^2$, we get,


$$=\int\frac{t^2+1}{1+t^4}dt + \int\frac{t^2-1}{1+t^4}dt$$ 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:


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


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good use of algebraic twins! – Ashish Gaurav Feb 23 '13 at 7:08

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