Equivalent $\int_{0}^{1}{x^2-x\over x^2+1}\cdot{\ln(x)\over x^3+1}dx=\int_{0}^{\infty}{x^2-x\over x^2+1}\cdot{\ln(1+x)\over x^3+1}\mathrm dx?$ Proposed:
Is $(1)$ equivalent to $(2)$ in term of transformation? 

$$\int_{0}^{1}{x^2-x\over x^2+1}\cdot{\ln(x)\over x^3+1}\mathrm dx={5\pi^2\over 2\cdot 6^3}\tag1$$

and

$$\int_{0}^{\infty}{x^2-x\over x^2+1}\cdot{\ln(1+x)\over x^3+1}\mathrm dx={5\pi^2\over 2\cdot 6^3}\tag2$$

 A: Let
$$I=\int_{0}^{1}{x^2-x\over x^2+1}\cdot{\ln(x)\over x^3+1}\mathrm dx$$
and
$$J=\int_{0}^{\infty}{x^2-x\over x^2+1}\cdot{\ln(1+x)\over x^3+1}\mathrm dx.$$
Note
\begin{eqnarray}
J&=&\int_{0}^{1}{x^2-x\over x^2+1}\cdot{\ln(1+x)\over x^3+1}\mathrm dx+\int_{1}^{\infty}{x^2-x\over x^2+1}\cdot{\ln(1+x)\over x^3+1}\mathrm dx\\
&=&\int_{0}^{1}{x^2-x\over x^2+1}\cdot{\ln(1+x)\over x^3+1}\mathrm dx+\int_{0}^{1}{\frac{1}{x^2}-\frac{1}{x}\over \frac{1}{x^2}+1}\cdot{\ln(1+\frac1x)\over \frac{1}{x^3}+1}\frac{1}{x^2}\mathrm dx\\
&=&\int_{0}^{1}{x^2-x\over x^2+1}\cdot{\ln(1+x)\over x^3+1}\mathrm dx+\int_{0}^{1}{x-x^2\over x^2+1}\cdot{\ln(\frac{1+x}x)\over x^3+1}\mathrm dx\\
&=&\int_{0}^{1}{x^2-x\over x^2+1}\cdot{\ln(x)\over x^3+1}\mathrm dx\\
&=&I
\end{eqnarray}
and
\begin{eqnarray}
I&=&\int_0^1\bigg[\frac{x^2}{1+x^3}-\frac{x}{1+x^2}\bigg]\ln x\mathrm dx\\
&=&\int_0^1\frac{x^2}{1+x^3}\ln x\mathrm dx-\int_0^1\frac{x}{1+x^2}\ln x\mathrm dx\\
&=&\frac19\int_0^1\frac{1}{1+x}\ln x\mathrm dx-\frac14\int_0^1\frac{1}{1+x}\ln x\mathrm dx\\
&=&-\frac{5}{36}\int_0^1\frac{1}{1+x}\ln x\mathrm dx\\
&=&-\frac{5}{36}(-\frac{\pi^2}{12})\\
&=&\frac{5\pi^2}{432}.
\end{eqnarray}
