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.
Best Answer
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 $$I=\int \frac{du}{u^2+2} +\int \frac{dv}{v^2-2}= \frac{1}{\sqrt{2}} \tan^{-1} \frac{u}{\sqrt{2}} + \frac{1}{2\sqrt{2}} \ln \frac{v-\sqrt{2}}{v+\sqrt{2}}=$$ $$ \frac{1}{\sqrt{2}} \tan^{-1}\frac{(t-1/t)}{\sqrt{2}}+\frac{1}{2 \sqrt{2}} \ln \frac{t+1/t-\sqrt{2}}{t+1/t+\sqrt{2}}, t=\sqrt{x}.$$