Calculating $\frac{1}{1\cdot 2\cdot 3}+\frac{1}{5\cdot 6\cdot 7}+\frac{1}{9\cdot 10\cdot 11}+\cdots$

We can write the general term of the series as

$$\frac{1}{(4k+1)(4k+2)(4k+3)}=\frac{1}{2}\left(\frac{1}{4k+1}-\frac{2}{4k+2}+\frac{1}{4k+3}\right)$$

Then, noting that $\int_0^1 x^{4k}\,dx=\frac{1}{4k+1}$, we have

$$\begin{align} \sum_{n=0}^\infty\frac{1}{2}\left(\frac{1}{4k+1}-\frac{2}{4k+2}+\frac{1}{4k+3}\right)&=\sum_{n=0}^\infty\frac12 \int_0^1 (x^{4k}-2x^{4k+1}+x^{4k+2})\,dx\\\\ &=\frac12 \int_0^1 \frac{(1-2x+x^2)}{1-x^4}\,dx\\\\ &=-\frac12\int_0^1 \frac{x-1}{(x^2+1)(x+1)}\,dx \end{align}$$

Can you finish now?

I have a suspicion that the following method would be more like the one expected of the candidates for this exam.

First we decompose into partial fractions, so, as given already, $$S=\frac 12\sum_{r=0}^{\infty}\left(\frac{1}{4k+1}-\frac{2}{4k+2}+\frac{1}{4k+3}\right)$$

Then we start by writing this out explicitly, so that $$2S=\left(\frac 11-\frac 22+\frac 13\right)+\left(\frac 15-\frac 26+\frac 17\right)+\left(\frac 19-\frac{2}{10}+\frac{1}{11}\right)+...$$

Then we systematically add in and subtract terms, so $$2S=\left(\frac 11-\frac 12+\frac 13-\frac 14\right)+\color{red}{\left(-\frac 12+\frac 14\right)}+\left(\frac 15-\frac 16+\frac 17-\frac 18\right)+\color{red}{\left(-\frac 16+\frac 18\right)}+\left(\frac 19-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}\right)+\color{red}{\left(-\frac{1}{10}+\frac{1}{12}\right)}+...$$

So $$2S=\ln 2-\color {red}{\frac 12\ln 2}$$

Then $$S=\frac 14\ln 2$$

I don't think the integration method as shown by @Dr. MV was expected to be known by those students...