Find the time it takes for a particle to reach the source of the attractive force acting on it A particle of mass m is released from rest a distance b from a fixed origin of force that attracts the particle according to the inverse square law:
$$F(x)=-kx^{-2}$$
Show that the time required for the particle to reach the origin is
$$t=\pi\sqrt{\frac{mb^3}{8k}}$$
My solution is as follows:
$$F(x)=-kx^{-2}\tag1$$
$$m\frac{d\dot x}{dt}=-kx^{-2}$$
By chain rule, $\frac{d\dot x}{dt}=\frac{dx}{dt}\frac{d\dot x}{dx}=\dot x\frac{d\dot x}{dx}$
$$m\dot{x}\frac{d\dot{x}}{dx}=-kx^{-2}\tag2$$
When the particle is at a distance $x$ from the source, wherein $x\leq b$, the speed of the particle is $\dot{x}$. Therefore,
$$\int_0^{\dot{x}}\dot{x}d\dot{x}=-\frac km\int_b^x\frac{dx}{x^2}$$
$$\frac12\dot{x}^2=\frac km\left(\frac1x-\frac1b\right)$$
Solving for $\dot x$, I get
$$\dot{x}=\sqrt{\frac{2k}{mb}\left(\frac{b-x}{x}\right)}\tag3$$
$$\frac{dx}{dt}=\sqrt{\frac{2k}{mb}\left(\frac{b-x}{x}\right)}$$
$$dt=\sqrt{\frac{mb}{2k}\left(\frac{x}{b-x}\right)}dx$$
When the particle has reached the source at time $t$, the distance to the source is $0$. Thus,
$$\int_0^tdt=\sqrt{\frac{mb}{2k}}\int_b^0\left(\frac{x}{b-x}\right)^\frac12dx\tag4$$
Applying partial fraction decomposition, I get
$$t=\sqrt{\frac{mb}{2k}}\left(b\int_b^0\frac{dx}{b-x}-\int_b^0dx\right)$$
Intergrating $\int_b^0\frac{dx}{b-x}$, I get $-\infty$. I don't know where to go from there. From the solution given to us, Eq. (4) was rewritten as
$$\int_0^tdt=\sqrt{\frac{mb^3}{2k}}\int_b^0\left(\frac{\frac xb}{1-\frac xb}\right)^\frac12d\left(\frac xb\right)\tag5$$
Since $x\leq b$, say $\frac xb=\sin^2\theta$
$$t=\sqrt{\frac{mb^3}{2k}}\int_{-\pi/2}^0\frac{\sin\theta\left(2\sin\theta\cos\theta\right)}{\cos\theta}d\theta\tag6$$
$$t=\sqrt{\frac{2mb^3}{k}}\int_{-\pi/2}^0\sin^2\theta d\theta$$
$$t=\pi\sqrt{\frac{mb^3}{8k}}$$
What I don't understand from the given solution is how could $dx$ be rewritten as $d\left(\frac xb\right)$ and how the limits for Eq. (6) were obtained. I'm also confused as to why trigonometric functions were used and by what reason was $\frac xb=\sin^2\theta$. From my solution, I know I'm wrong, but I'd like to know if there are other integration methods that could be used for Eq. (6) without using trigonometric functions.
 A: (1)
There is a minor issue here. Since in your case, $t$ increases as $x$ decreases, you should get 
$$dt=-\sqrt{\frac{mb}{2k}\left(\frac{x}{b-x}\right)}dx.$$
Therefore, the required time would be
$$\sqrt{\frac{mb}{2k}}\int_0^b
\sqrt{\frac{x}{b-x}}dx.$$
(2)
I don’t think it is obvious to introduce trigonometric function by directly observing the integrand here. So let me present it in another way. 
Use the substitution $u=\sqrt{\frac{x}{b-x}}$, we have $x=b\left(1-\frac{1}{u^2+1}\right)$ and we can replace $\int_0^b\sqrt{\frac{x}{b-x}}dx$ by
$$ \int_0^{\infty} 
u\frac{2bu}{(u^2+1)^2}du
=2b\int_0^{\infty} \frac{u^2}{(u^2+1)^2}du.$$
Now it is a standard form where we have to use the substitution $u=\tan{\theta}$, and get
$$ 2b \int_0^{\frac{\pi}{2}} \frac{\tan^2\theta}{\sec^2\theta}d\theta = 2b\times \frac{\pi}{4},$$
which gives the result. 
Remark If you insist not to use trigonometric function to find the anti-derivative of $\frac{u^2}{(u^2+1)^2}$, you could use the integration by part on the decomposition
$$ u\times \frac{u}{(u^2+1)^2}du, $$
where $U=u$ and $dV=\frac{u}{(u^2+1)^2}du$. 
