A mass starts from rest at a distance a from the centre of force which attracts inversely as the distance. Find the time to reach the centre
Here acceleration is inversely proportional to the distance and initial velocity is zero. But the acceleration keeps on increasing and not constant.
So how to proceed
Let us say your center (the force emitter) is the origin $0$ of the real line and the particle (the receiver) is at $x(0)=a$ where $a>0$. You are asked to find $t$ such that $x(t)=0$, given that $x''(t)=-1/x(t)$, that is, the acceleration is inversely proportional to the distance (well, you didn't say what is the coefficient of proportionality, so I've chosen it to be $-1$ for the particle to move towards the origin) and $x'(0)=0$, that is, the initial velocity is zero.
Solve \begin{cases} a>0,\\t\geq0,\\x(0)=a,\\x'(0)=0,\\\forall_t x''(t)=-1/x(t),\\x(t)=0; \end{cases} for $t$. For $x''(t)=-1/x(t)$, multiply both sides by $x'(t)$: $$x''(t)x'(t)=-\frac{x'(t)}{x(t)};$$ integrate both sides with respect to $t$: $$\int x''(t)x'(t)\mathrm dt=-\int \frac{x'(t)}{x(t)}\mathrm dt;$$ for the left-hand side integral, substitute $u=x'(t)\implies \mathrm du=x''(t)dt$: $$\int u\mathrm du=-\int \frac{x'(t)}{x(t)}\mathrm dt;$$ the antiderivative of $u$ with respect to $u$ is $\dfrac{u^2}{2}$: $$\frac{u^2}{2}=-\int \frac{x'(t)}{x(t)}\mathrm dt;$$ multiply both sides by $2$: $$u^2=-2\int \frac{x'(t)}{x(t)}\mathrm dt;$$ for the right-hand side integral, substitute $s=x(t)\implies\mathrm ds=x'(t)\mathrm dt$: $$u^2=-2\int \frac{\mathrm ds}{s};$$ the integral of $-2\mathrm ds/s$ is $-2\log(s)+c_1$, where $c_1$ is an arbitrary constant: $$u^2=-2\log(s)+c_1;$$ substitute back for $u$: $$x'(t)^2=-2\log(s)+c_1;$$ substitute back for $s$: $$x'(t)^2=-2\log(x(t))+c_1;$$ from the initial condition $x'(0)=0$, solve for $c_1$: $$0=-2\log(x(0))+c_1\iff c_1=2\log(x(0));$$ from the initial condition $x(0)=a$, substitute for $x(0)$: $$c_1=2\log(a);$$ substitute for $c_1$: $$x'(t)^2=-2\log(x(t))+2\log(a)=2\log(a/x(t));$$ take the square roots of both sides: $$x'(t)=\pm\sqrt{2\log(a/x(t))};$$ divide both sides by $\sqrt{2\log(a/x(t))}$: $$\frac{x'(t)}{\sqrt{2\log(a/x(t))}}=\pm 1;$$ integrate both sides with respect to $t$: $$\int\frac{x'(t)}{\sqrt{2\log(a/x(t))}}\mathrm dt=\pm\int\mathrm dt;$$ the integral of $\mathrm dt$ is $t$: $$\int\frac{x'(t)}{\sqrt{2\log(a/x(t))}}\mathrm dt=\pm t;$$ substitute $r=x(t)\implies\mathrm dr=x'(t)\mathrm dt$: $$\int\frac{\mathrm dr}{\sqrt{2\log(a/r)}}=\pm t;$$ multiply both sides by $\sqrt{2}/a$: $$\int\frac{\mathrm dr/a}{\sqrt{\log(a/r)}}=\pm t\frac{\sqrt{2}}{a};$$ substitute $q=r/a\implies\mathrm dq=\mathrm dr/a$: $$\int\frac{\mathrm dq}{\sqrt{\log(1/q)}}=\pm t\frac{\sqrt{2}}{a};$$ substitute $y=-\log(q)\implies q=e^{-y}\implies\mathrm dq=-e^{-y}\mathrm dy$: $$-\int\frac{e^{-y}}{\sqrt{\log(1/e^{-y})}}\mathrm dy=\pm t\frac{\sqrt{2}}{a};$$ rewrite $\log(1/e^{-y})=\log(e^y)=y$: $$-\int\frac{e^{-y}}{\sqrt{y}}\mathrm dy=\pm t\frac{\sqrt{2}}{a};$$ substitute $w=\sqrt{y}\implies y=w^2\implies\mathrm dy=2w\mathrm dw$: $$-\int\frac{e^{-w^2}}{w}2w\mathrm dw=\pm t\frac{\sqrt{2}}{a};$$ rewrite $\displaystyle\int\dfrac{e^{-w^2}}{w}2w\mathrm dw=2\int e^{-w^2}\mathrm dw$: $$-2\int e^{-w^2}\mathrm dw=\pm t\frac{\sqrt{2}}{a};$$ the integral of $e^{-w^2}\mathrm dw$ is $\dfrac{\sqrt{\pi}}{2}\operatorname{erf}(w)+c_2$, where $c_2$ is an arbitrary constant: $$-2\frac{\sqrt{\pi}}{2}\operatorname{erf}(w)+c_2=\pm t\frac{\sqrt{2}}{a};$$ rewrite $2\frac{\sqrt{\pi}}{2}=\sqrt{\pi}$: $$-\sqrt{\pi}\operatorname{erf}(w)+c_2=\pm t\frac{\sqrt{2}}{a};$$ substitute back for $w$: $$-\sqrt{\pi}\operatorname{erf}\left(\sqrt{y}\right)+c_2=\pm t\frac{\sqrt{2}}{a};$$ substitute back for $y$: $$-\sqrt{\pi}\operatorname{erf}\left(\sqrt{-\log\left(q\right)}\right)+c_2=\pm t\frac{\sqrt{2}}{a};$$ substitute back for $q$: $$-\sqrt{\pi}\operatorname{erf}\left(\sqrt{-\log\left(r/a\right)}\right)+c_2=\pm t\frac{\sqrt{2}}{a};$$ substitute back for $r$: $$-\sqrt{\pi}\operatorname{erf}\left(\sqrt{-\log\left(x\left(t\right)/a\right)}\right)+c_2=\pm t\frac{\sqrt{2}}{a};$$ from the initial condition $x(0)=a$, solve for $c_2$: $$-\sqrt{\pi}\operatorname{erf}\left(\sqrt{-\log\left(a/a\right)}\right)+c_2=0\iff-\sqrt{\pi}\operatorname{erf}(0)+c_2=0\iff c_2=0;$$ substitute back for $c_2$: $$-\sqrt{\pi}\operatorname{erf}\left(\sqrt{-\log\left(x\left(t\right)/a\right)}\right)=\pm t\frac{\sqrt{2}}{a};$$ take the limit as $x(t)\to 0$: $$\pm t\frac{\sqrt{2}}{a}=-\sqrt{\pi}\operatorname{erf}\left(\sqrt{-\log\left(x\left(t\right)/a\right)}\right)=-\sqrt{\pi}\operatorname{erf}\left(\sqrt{\infty}\right)=-\sqrt{\pi}\operatorname{erf}\left(\infty\right)=-\sqrt{\pi};$$ multiply both sides by $\pm a/\sqrt{2}$: $$t=\mp a\frac{\sqrt{\pi}}{2};$$ from the initial condition $t\geq0$, select the positive solution and write out the answer: $$\therefore t=\boxed{a\frac{\sqrt{\pi}}{2}}\phantom{.}.$$