Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
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
add comment

2 Answers 2

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

share|improve this answer
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
add comment


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


share|improve this answer
good use of algebraic twins! –  Ashish Gaurav Feb 23 '13 at 7:08
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.