I'm trying to find the antiderivative of

$$ I = \int \frac{dx}{\sqrt{x} (1 + x^2)}$$

but I've been stuck on it for a while. (I came across it in this Youtube video).

I know the antiderivative of $\frac{1}{1 + x^2}$ is $\arctan x$, but I'm not sure that's relevant due to the extra factor. I also know that the antiderivative of $\frac{1}{\sqrt{x}}$ is $2x^{1/2}$, but again, I'm not sure that's relevant.

I can re-write the expression in a few alternate forms, but I don't know how to proceed from these alternate forms either.

$$\begin{aligned} I &= \int \frac{dx}{\sqrt{x} + x^2 \sqrt{x}} \\ I &= \int \left[\frac{1}{\sqrt{x}} \cdot \frac{1}{1 + x^2} \right] dx \end{aligned}$$

I learned how to integrate rational functions, but only when the numerator could be re-stated as different linear terms, and it doesn't seem to be possible here.

I also tried integrating by substitution and that didn't seem to be going anywhere either.

I think I must be missing something. Can anyone help?

Update 1: Multiple answers and comments below suggested using substitution, which I had actually done in my notes. I managed to obtain:

$$ 2 \int \frac{du}{(1 + t^4)}$$

But it appears my problem was the partial fraction decomposition step. Maybe I'm not applying it correctly.

Someone suggested using the fact that $1 + t^4 = (t^2 - \sqrt{2}t + 1)(t^2 + \sqrt{2}t + 1)$.

In that case, if I'm not mistaken, I must find values $A$ and $B$ such that

$$ \frac{2}{(t^2 - \sqrt{2}t + 1)(t^2 + \sqrt{2}t + 1)} = \frac{A}{(t^2 - \sqrt{2}t + 1)} + \frac{B}{(t^2 + \sqrt{2}t + 1)}$$

and it must be the case that

$$ 2 = t^2(A+B) + \sqrt{2} t (A-B) + (A+B) $$

But when I've done partial fraction decomposition in the past, the left-hand-side in the last line would have had three terms of matching degrees, which is not the case this time.

Am I doing something wrong?

I've also studied the more detailed answer provided below, but wasn't able to follow it all the way. It might be using techniques I'm not familiar with yet.


Let $x=t^2$ and write $$I=2 \int \frac{dt}{1+t^4}=\int \frac{(1+t^2)+(1-t^2)}{1+t^4}dt =\int \frac{(1+1/t^2)dt}{t^2+1/t^2}+ \int \frac{(1-1/t^2)dt}{t^2+1/t^2}~~~~(1)$$ $$\Rightarrow I= \int \frac{(1+1/t^2)dt}{(t-1/t)^2+2} + \frac{(1-1/t^2)dt}{(t+1/t)^2-2} ~~~(2)$$ Use $u=t-1/t$ and $v=t+1/t$ in the first and second integrals, respectively. Then $(1-1/t^2)dt=du$, $(1+1/t^2)dt=dv$ and $$I=\int \frac{du}{u^2+2} +\int \frac{dv}{v^2-2}= \frac{1}{\sqrt{2}} \tan^{-1} \frac{u}{\sqrt{2}} + \frac{1}{2\sqrt{2}} \ln \frac{v-\sqrt{2}}{v+\sqrt{2}}=$$ $$ \frac{1}{\sqrt{2}} \tan^{-1}\frac{(t-1/t)}{\sqrt{2}}+\frac{1}{2 \sqrt{2}} \ln \frac{t+1/t-\sqrt{2}}{t+1/t+\sqrt{2}}, t=\sqrt{x}.$$

  • $\begingroup$ This problem and solution is very old and I've seen it a lot on many Facebook groups. We would like to see new challenging problems. $\endgroup$ – Ali Shather Aug 12 at 6:05
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    $\begingroup$ But Sir one should answer in any case Similarly, one may think that A,B,C,... and 1,2,3,...should not be taught to the kids any more because we all know it. Also now I understantd as to why other answers here avoid this old stuff! $\endgroup$ – Dr Zafar Ahmed DSc Aug 12 at 6:43
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    $\begingroup$ Dr. Zafar, thank you for this answer. It is very helpful and I'm still studying it. The problem is new to me, and I'm not part of the many Facebook groups that @Ali Shather has seen them on. Furthermore, what is new and challenging is subjective to every person's experience. Something that might be new and challenging to Ali Shather might be old hat to someone else. Mathematics Stack Exchange is for people studying mathematics at any level (math.stackexchange.com/help/on-topic) so my question and your answer are entirely on topic for the site. $\endgroup$ – Calculemus Aug 12 at 6:55
  • $\begingroup$ @calculemus I am going to edit it to make it clearer. $\endgroup$ – Dr Zafar Ahmed DSc Aug 12 at 7:42
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    $\begingroup$ @calculemus I have edited it again. In Eq. (1) it just an algerbraic decomposition of $2=1+t^2+1-t^2$, next we divide by $t^2$ up and down.. In Num we express $t^2+1/t^2$ in two ways. Twin transformations $u,v$ have been used. the $u$ and $v$ integrals are standard ones. $\endgroup$ – Dr Zafar Ahmed DSc Aug 12 at 9:13

If, as suggested in comments, to use $$t=\sqrt x \implies x=t^2 \implies dx=2t\,dt$$ you should end with $$I = \int \frac{dx}{\sqrt{x} (1 + x^2)}=2\int\frac{dt}{1+t^4}$$ and use the fact that $$t^4+1=\left(t^2-\sqrt{2} t+1\right) \left(t^2+\sqrt{2} t+1\right)$$ Now, partial fraction decomposition as Parcly Taxel wrote while I was typing.


Under $x=t^2$, one has \begin{eqnarray} I &=& \int \frac{dx}{\sqrt{x} (1 + x^2)}=2\int\frac{dt}{1+t^4}\\ &=&2\int\frac{1}{t^2+\frac{1}{t^2}}\frac{dt}{t^2}=\int\frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}}dt+\int\frac{1-\frac{1}{t^2}}{t^2+\frac{1}{t^2}}dt\\ &=&\int\frac{1}{(t-\frac{1}{t})^2+2}d(t-\frac1t)+\int\frac{1}{(t+\frac{1}{t})^2-2}d(t+\frac1t)\\ &=&\frac1{\sqrt2}\arctan(\frac{t-\frac1t}{\sqrt2})+\frac1{\sqrt2}\text{arctanh}(\frac{t+\frac1t}{\sqrt2})\\ &=&\frac1{\sqrt2}\arctan(\frac{\sqrt x-\frac1{\sqrt x}}{\sqrt2})-\frac1{\sqrt2}\text{arctanh}(\frac{\sqrt x+\frac1{\sqrt x}}{\sqrt2}). \end{eqnarray}

  • $\begingroup$ In the last step $t=\sqrt{x}$, instead of $x^2$/ $\endgroup$ – Dr Zafar Ahmed DSc Aug 12 at 17:21
  • $\begingroup$ @DrZafarAhmedDSc, oh, yes. Thanks. $\endgroup$ – xpaul Aug 13 at 12:47

Let's try the substitution in comments: $$\int\frac1{\sqrt x(1+x^2)}\,dx=2\int\frac1{1+u^4}\,du$$ The $2$ appears because $\frac{du}{dx}=\frac1{2\sqrt x}$. Now a rational function is left, which can be integrated via partial fraction decomposition.


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