Let $\mathcal{I}$ denote the value of the following logarithmic integral:
$$\mathcal{I}:=\int_{0}^{1}\mathrm{d}x\,\frac{3x^{2}}{1+x^{3}}\ln{\left(1-x^{4}\right)}\approx-0.47159.$$
It will be shown that
$$\mathcal{I}=-\frac12\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{5\pi^{2}}{144}+\frac34\ln^{2}{\left(2\right)}.$$
Making use of the factorizations $\left(1+x^{3}\right)=\left(1+x\right)\left(1-x+x^{2}\right)$ and $\left(1-x^{4}\right)=\left(1-x\right)\left(1+x\right)\left(1+x^{2}\right)$, we can split up the integral into simpler components as
$$\begin{align}
\mathcal{I}
&=\int_{0}^{1}\mathrm{d}x\,\frac{3x^{2}}{1+x^{3}}\ln{\left(1-x^{4}\right)}\\
&=\int_{0}^{1}\mathrm{d}x\,\left[\frac{1}{1+x}+\frac{2x-1}{1-x+x^{2}}\right]\left[\ln{\left(1-x\right)}+\ln{\left(1+x\right)}+\ln{\left(1+x^{2}\right)}\right]\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{1}{1+x}\left[\ln{\left(1-x\right)}+\ln{\left(1+x\right)}+\ln{\left(1+x^{2}\right)}\right]\\
&~~~~~+\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\left[\ln{\left(1-x\right)}+\ln{\left(1+x\right)}+\ln{\left(1+x^{2}\right)}\right]\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1-x\right)}}{1+x}+\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+x\right)}}{1+x}+\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+x^{2}\right)}}{1+x}\\
&~~~~~+\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1-x\right)}+\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1+x\right)}\\
&~~~~~+\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1+x^{2}\right)}\\
&=:\mathcal{I}_{1}+\mathcal{I}_{2}+\mathcal{I}_{3}+\mathcal{I}_{4}+\mathcal{I}_{5}+\mathcal{I}_{6}.\\
\end{align}$$
Let's focus on the last integral first.
Using the derivatives
$$\frac{d}{dx}\ln{\left(1-x+x^{2}\right)}=\frac{2x-1}{1-x+x^{2}},$$
and
$$\frac{d}{dx}\frac{\ln^{2}{\left(1-x+x^{2}\right)}}{2}=\frac{2x-1}{1-x+x^{2}}\ln{\left(1-x+x^{2}\right)},$$
we show that the following integral vanishes identically:
$$\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1-x+x^{2}\right)}=\frac{\ln^{2}{\left(1-x+x^{2}\right)}}{2}\bigg{|}_{0}^{1}=0.$$
Then, $\mathcal{I}_{6}$ can be rewritten in the following way:
$$\begin{align}
\mathcal{I}_{6}
&=\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1+x^{2}\right)}\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1+x^{2}\right)}-\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1-x+x^{2}\right)}\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\left[\ln{\left(1+x^{2}\right)}-\ln{\left(1-x+x^{2}\right)}\right]\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(\frac{1+x^{2}}{1-x+x^{2}}\right)}\\
&=\int_{1}^{0}\mathrm{d}y\,\frac{\left(-2\right)}{\left(1+y\right)^{2}}\cdot\frac{\left(1+y\right)\left(1-3y\right)}{\left(1+3y^{2}\right)}\ln{\left(\frac{2+2y^{2}}{1+3y^{2}}\right)};~~~\small{\left[x=\frac{1-y}{1+y}\right]}\\
&=\int_{0}^{1}\mathrm{d}y\,\frac{2}{\left(1+y\right)}\cdot\frac{\left(1-3y\right)}{\left(1+3y^{2}\right)}\ln{\left(\frac{2+2y^{2}}{1+3y^{2}}\right)}\\
&=\int_{0}^{1}\mathrm{d}y\,\left[\frac{2}{\left(1+y\right)}-\frac{6y}{\left(1+3y^{2}\right)}\right]\ln{\left(\frac{2+2y^{2}}{1+3y^{2}}\right)}\\
&=\int_{0}^{1}\mathrm{d}y\,\frac{2}{\left(1+y\right)}\ln{\left(\frac{2+2y^{2}}{1+3y^{2}}\right)}-\int_{0}^{1}\mathrm{d}y\,\frac{6y}{\left(1+3y^{2}\right)}\ln{\left(\frac{2+2y^{2}}{1+3y^{2}}\right)}\\
&=\int_{0}^{1}\mathrm{d}y\,\frac{2}{\left(1+y\right)}\left[\ln{\left(2\right)}+\ln{\left(1+y^{2}\right)}-\ln{\left(1+3y^{2}\right)}\right]\\
&~~~~~-\int_{0}^{1}\mathrm{d}t\,\frac{3}{\left(1+3t\right)}\ln{\left(\frac{2+2t}{1+3t}\right)};~~~\small{\left[y^{2}=t\right]}\\
&=2\ln{\left(2\right)}\int_{0}^{1}\mathrm{d}y\,\frac{1}{\left(1+y\right)}+2\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y^{2}\right)}}{\left(1+y\right)}-2\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+3y^{2}\right)}}{\left(1+y\right)}\\
&~~~~~-\int_{0}^{1}\mathrm{d}u\,\frac{3}{\left(1+3u\right)}\ln{\left(1+u\right)};~~~\small{\left[t=\frac{1-u}{1+3u}\right]}\\
&=2\ln^{2}{\left(2\right)}+2\mathcal{I}_{3}-2\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+3y^{2}\right)}}{\left(1+y\right)}\\
&~~~~~+\operatorname{Li}_{2}{\left(\frac34\right)}-\operatorname{Li}_{2}{\left(\frac12\right)}+\operatorname{Li}_{2}{\left(-1\right)}\\
&=\operatorname{Li}_{2}{\left(\frac34\right)}-\operatorname{Li}_{2}{\left(\frac12\right)}+\operatorname{Li}_{2}{\left(-1\right)}+2\ln^{2}{\left(2\right)}+2\mathcal{I}_{3}\\
&~~~~~-2\int_{1}^{2}\mathrm{d}t\,\frac{\ln{\left(1+3(t-1)^{2}\right)}}{t};~~~\small{\left[1+y=t\right]}\\
&=\operatorname{Li}_{2}{\left(\frac34\right)}-\operatorname{Li}_{2}{\left(\frac12\right)}+\operatorname{Li}_{2}{\left(-1\right)}+2\ln^{2}{\left(2\right)}+2\mathcal{I}_{3}\\
&~~~~~-2\int_{1}^{2}\mathrm{d}t\,\frac{\ln{\left(3t^{2}-6t+4\right)}}{t}\\
&=\operatorname{Li}_{2}{\left(\frac34\right)}-\operatorname{Li}_{2}{\left(\frac12\right)}+\operatorname{Li}_{2}{\left(-1\right)}+2\ln^{2}{\left(2\right)}+2\mathcal{I}_{3}\\
&~~~~~-2\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}u\,\frac{\ln{\left(4u^{2}-4\sqrt{3}\,u+4\right)}}{u};~~~\small{\left[t=\frac{2u}{\sqrt{3}}\right]}\\
&=\operatorname{Li}_{2}{\left(\frac34\right)}-\operatorname{Li}_{2}{\left(\frac12\right)}+\operatorname{Li}_{2}{\left(-1\right)}+2\ln^{2}{\left(2\right)}+2\mathcal{I}_{3}\\
&~~~~~-2\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}u\,\frac{\ln{\left(4\right)}}{u}-2\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}u\,\frac{\ln{\left(u^{2}-\sqrt{3}\,u+1\right)}}{u}\\
&=2\mathcal{I}_{3}-\frac{\pi^{2}}{6}-\frac32\ln^{2}{\left(2\right)}+\operatorname{Li}_{2}{\left(\frac34\right)}-2\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{3}\,x+x^{2}\right)}}{x}.\\
\end{align}$$
The integral $\mathcal{I}_{3}$ can in turn be reduced to
$$\begin{align}
\mathcal{I}_{3}
&=\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+x^{2}\right)}}{1+x}\\
&=\int_{1}^{2}\mathrm{d}y\,\frac{\ln{\left(1+(y-1)^{2}\right)}}{y};~~~\small{\left[1+x=y\right]}\\
&=\int_{1}^{2}\mathrm{d}y\,\frac{\ln{\left(2-2y+y^{2}\right)}}{y}\\
&=\int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}}\mathrm{d}t\,\frac{\ln{\left(2-2t\sqrt{2}+2t^{2}\right)}}{t};~~~\small{\left[y=t\sqrt{2}\right]}\\
&=\int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}}\mathrm{d}t\,\frac{\ln{\left(2\right)}}{t}+\int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}}\mathrm{d}t\,\frac{\ln{\left(1-\sqrt{2}\,t+t^{2}\right)}}{t}\\
&=\ln^{2}{\left(2\right)}+\int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{2}\,x+x^{2}\right)}}{x}.\\
\end{align}$$
Next up, $\mathcal{I}_{5}$ can be reduced to a similar expression involving the same undetermined integral in the expression for $\mathcal{I}_{6}$:
$$\begin{align}
\mathcal{I}_{5}
&=\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1+x\right)}\\
&=-\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1-x+x^{2}\right)}}{1+x};~~~\small{I.B.P.s}\\
&=-\int_{1}^{2}\mathrm{d}y\,\frac{\ln{\left(3-3y+y^{2}\right)}}{y};~~~\small{\left[x=y-1\right]}\\
&=-\int_{\frac{1}{\sqrt{3}}}^{\frac{2}{\sqrt{3}}}\mathrm{d}t\,\frac{\ln{\left(3-3t\sqrt{3}+3t^{2}\right)}}{t};~~~\small{\left[y=t\sqrt{3}\right]}\\
&=-\int_{\frac{1}{\sqrt{3}}}^{\frac{2}{\sqrt{3}}}\mathrm{d}t\,\frac{\ln{\left(3\right)}}{t}-\int_{\frac{1}{\sqrt{3}}}^{\frac{2}{\sqrt{3}}}\mathrm{d}t\,\frac{\ln{\left(1-\sqrt{3}\,t+t^{2}\right)}}{t}\\
&=-\ln{\left(2\right)}\ln{\left(3\right)}-\int_{\frac{1}{\sqrt{3}}}^{\frac{2}{\sqrt{3}}}\mathrm{d}t\,\frac{\ln{\left(1-\sqrt{3}\,t+t^{2}\right)}}{t}\\
&=-\ln{\left(2\right)}\ln{\left(3\right)}-\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}u\,\frac{\ln{\left(\frac{1-\sqrt{3}\,u+u^{2}}{u^{2}}\right)}}{u};~~~\small{\left[t=u^{-1}\right]}\\
&=-\ln{\left(2\right)}\ln{\left(3\right)}+\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}u\,\frac{2\ln{\left(u\right)}-\ln{\left(1-\sqrt{3}\,u+u^{2}\right)}}{u}\\
&=-\ln{\left(2\right)}\ln{\left(3\right)}+\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}u\,\frac{2\ln{\left(u\right)}}{u}-\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}u\,\frac{\ln{\left(1-\sqrt{3}\,u+u^{2}\right)}}{u}\\
&=-\ln^{2}{\left(2\right)}-\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{3}\,x+x^{2}\right)}}{x}.\\
\end{align}$$
The remaining three integrals $\mathcal{I}_{1,2,4}$ each have nice exact values in terms of well-known constants: we have
$$\mathcal{I}_{1}=\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1-x\right)}}{1+x}=-\operatorname{Li}_{2}{\left(\frac12\right)}=\frac12\ln^{2}{\left(2\right)}-\frac{\pi^{2}}{12},$$
$$\mathcal{I}_{2}=\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+x\right)}}{1+x}=\frac12\ln^{2}{\left(2\right)},$$
and
$$\begin{align}
\mathcal{I}_{4}
&=\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1-x\right)}\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1-x+x^{2}\right)}}{1-x};~~~\small{I.B.P.s}\\
&=\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1-y+y^{2}\right)}}{y};~~~\small{\left[x=1-y\right]}\\
&=\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y^{3}\right)}}{y}-\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}}{y}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1+t\right)}}{3t};~~~\small{\left[y^{3}=t\right]}\\
&~~~~~-\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}}{y}\\
&=-\frac23\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1+t\right)}}{t}\\
&=\frac23\operatorname{Li}_{2}{\left(-1\right)}\\
&=-\frac{\pi^{2}}{18}.\\
\end{align}$$
Then,
$$\begin{align}
\mathcal{I}
&=\mathcal{I}_{1}+\mathcal{I}_{2}+\mathcal{I}_{3}+\mathcal{I}_{4}+\mathcal{I}_{5}+\mathcal{I}_{6}\\
&=\left[\frac12\ln^{2}{\left(2\right)}-\frac{\pi^{2}}{12}\right]+\frac12\ln^{2}{\left(2\right)}+\mathcal{I}_{3}-\frac{\pi^{2}}{18}\\
&~~~~~-\ln^{2}{\left(2\right)}-\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{3}\,x+x^{2}\right)}}{x}\\
&~~~~~+2\mathcal{I}_{3}-\frac{\pi^{2}}{6}-\frac32\ln^{2}{\left(2\right)}+\operatorname{Li}_{2}{\left(\frac34\right)}-2\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{3}\,x+x^{2}\right)}}{x}\\
&=\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{11\pi^{2}}{36}-\frac32\ln^{2}{\left(2\right)}+3\mathcal{I}_{3}\\
&~~~~~-3\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{3}\,x+x^{2}\right)}}{x}\\
&=\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{11\pi^{2}}{36}-\frac32\ln^{2}{\left(2\right)}\\
&~~~~~+3\left[\ln^{2}{\left(2\right)}+\int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{2}\,x+x^{2}\right)}}{x}\right]\\
&~~~~~-3\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{3}\,x+x^{2}\right)}}{x}\\
&=\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{11\pi^{2}}{36}+\frac32\ln^{2}{\left(2\right)}\\
&~~~~~+3\int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{2}\,x+x^{2}\right)}}{x}-3\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{3}\,x+x^{2}\right)}}{x}.\\
\end{align}$$
At this point, it will be helpful to introduce a two-variable variant of the dilogarithm function:
$$\operatorname{Li}_{2}{\left(r,\theta\right)}:=-\frac12\int_{0}^{r}\mathrm{d}x\,\frac{\ln{\left(1-2x\cos{\left(\theta\right)}+x^{2}\right)}}{x};~~~\small{\left(r,\theta\right)\in\mathbb{R}^{2}}.$$
This function can be shown to satisfy the following pair of formulas useful to the problem at hand:
$$\operatorname{Li}_{2}{\left(\cos{\left(\theta\right)},\theta\right)}=\frac12\left(\frac{\pi}{2}-\theta\right)^{2}+\frac14\operatorname{Li}_{2}{\left(\cos^{2}{\left(\theta\right)}\right)};~~~\small{\theta\in\left[0,\pi\right]},$$
$$\operatorname{Li}_{2}{\left(2\cos{\left(\theta\right)},\theta\right)}=\left(\frac{\pi}{2}-\theta\right)^{2};~~~\small{\theta\in\left[0,\pi\right]}.$$
Then, for $\theta\in\left(0,\frac{\pi}{2}\right)$, we have
$$\begin{align}
\int_{\cos{\left(\theta\right)}}^{2\cos{\left(\theta\right)}}\mathrm{d}x\,\frac{\ln{\left(1-2x\cos{\left(\theta\right)}+x^{2}\right)}}{x}
&=\int_{0}^{2\cos{\left(\theta\right)}}\mathrm{d}x\,\frac{\ln{\left(1-2x\cos{\left(\theta\right)}+x^{2}\right)}}{x}\\
&~~~~~-\int_{0}^{\cos{\left(\theta\right)}}\mathrm{d}x\,\frac{\ln{\left(1-2x\cos{\left(\theta\right)}+x^{2}\right)}}{x}\\
&=-2\operatorname{Li}_{2}{\left(2\cos{\left(\theta\right)},\theta\right)}+2\operatorname{Li}_{2}{\left(\cos{\left(\theta\right)},\theta\right)}\\
&=-2\left(\frac{\pi}{2}-\theta\right)^{2}+2\left[\frac12\left(\frac{\pi}{2}-\theta\right)^{2}+\frac14\operatorname{Li}_{2}{\left(\cos^{2}{\left(\theta\right)}\right)}\right]\\
&=\frac12\operatorname{Li}_{2}{\left(\cos^{2}{\left(\theta\right)}\right)}-\left(\frac{\pi}{2}-\theta\right)^{2}.\\
\end{align}$$
We can now complete our calculation for $\mathcal{I}$:
$$\begin{align}
\mathcal{I}
&=\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{11\pi^{2}}{36}+\frac32\ln^{2}{\left(2\right)}\\
&~~~~~+3\int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{2}\,x+x^{2}\right)}}{x}-3\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{3}\,x+x^{2}\right)}}{x}\\
&=\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{11\pi^{2}}{36}+\frac32\ln^{2}{\left(2\right)}\\
&~~~~~+3\int_{\cos{\left(\frac{\pi}{4}\right)}}^{2\cos{\left(\frac{\pi}{4}\right)}}\mathrm{d}x\,\frac{\ln{\left(1-2x\cos{\left(\frac{\pi}{4}\right)}+x^{2}\right)}}{x}\\
&~~~~~-3\int_{\cos{\left(\frac{\pi}{6}\right)}}^{2\cos{\left(\frac{\pi}{6}\right)}}\mathrm{d}x\,\frac{\ln{\left(1-2x\cos{\left(\frac{\pi}{6}\right)}+x^{2}\right)}}{x}\\
&=\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{11\pi^{2}}{36}+\frac32\ln^{2}{\left(2\right)}\\
&~~~~~+3\left[\frac12\operatorname{Li}_{2}{\left(\cos^{2}{\left(\frac{\pi}{4}\right)}\right)}-\left(\frac{\pi}{2}-\frac{\pi}{4}\right)^{2}\right]\\
&~~~~~-3\left[\frac12\operatorname{Li}_{2}{\left(\cos^{2}{\left(\frac{\pi}{6}\right)}\right)}-\left(\frac{\pi}{2}-\frac{\pi}{6}\right)^{2}\right]\\
&=\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{11\pi^{2}}{36}+\frac32\ln^{2}{\left(2\right)}\\
&~~~~~+\frac32\operatorname{Li}_{2}{\left(\frac12\right)}-\frac{3\pi^{2}}{16}\\
&~~~~~-\frac32\operatorname{Li}_{2}{\left(\frac34\right)}+\frac{\pi^{2}}{3}\\
&=-\frac12\operatorname{Li}_{2}{\left(\frac34\right)}+\frac{\pi^{2}}{36}+\frac32\ln^{2}{\left(2\right)}\\
&~~~~~+\frac32\left[\frac{\pi^{2}}{12}-\frac12\ln^{2}{\left(2\right)}\right]-\frac{3\pi^{2}}{16}\\
&=-\frac12\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{5\pi^{2}}{144}+\frac34\ln^{2}{\left(2\right)}.\blacksquare\\
\end{align}$$
Best Answer
Let us try "converting it to a double summation". $$\begin{aligned}S&=\sum_{t=1}^{37}\sum_{j=t}^{37}\frac tj\\ &=\sum_{j=1}^{37}\sum_{t=1}^{j}\frac tj\\ &=\sum_{j=1}^{37}\frac{j(j+1)}{2j}\\ &=\sum_{j=1}^{37}\frac{j+1}2\\ &=\frac{37\cdot40}{2\cdot2}\\ &=370 \end{aligned}$$