When we plug in the t-distribution into the definition of the entropy we have to do an integral of the form
$$H=-C_{\nu}\int_{-\infty}^{\infty}\left(1+\frac{t^2}{\nu}\right)^{-\frac{1+\nu}{2}}\left[\ln C_{\nu}-\frac{\nu+1}{2}\ln\left(1+\frac{t^2}{\nu}\right)\right]=H_1+H_2$$
where the normalization constant is denoted by $C_{\nu}^{-1}=\sqrt{\nu}B\left(\frac{1}{2},\frac{\nu}{2}\right)$. The integrals $H_1$ and $H_2$ denote the integrals generated by the two terms in the brackets respectively. For $H_1$ it suffices to note that it is proportional to the normalization condition of the function so we easily get
$$H_1=\ln C_{\nu}^{-1}$$
For the second term we perform a change of variables $x=\frac{t^2}{\nu}$:
$$H_2=\frac{\nu+1}{2}\sqrt{\nu}C_{\nu}\int_{0}^{\infty}(1+x)^{-\frac{1+\nu}{2}}\ln(1+x)\frac{dx}{\sqrt{x}}$$
This integral can be written in terms of the Beta function
$$\int_{0}^{\infty}(1+x)^{-\frac{1+\nu}{2}}\ln(1+x)\frac{dx}{\sqrt{x}}=-2\frac{d}{d\nu}\int_{0}^{\infty}dx\frac{x^{-1/2}}{(1+x)^{\frac{\nu+1}{2}}}=-2\frac{dB\left(\frac{1}{2},\frac{\nu}{2}\right)}{d\nu}
$$
So the second integral is equal to
$$H_2=-(\nu+1)\frac{d}{d\nu}\ln B\left(\frac{1}{2},\frac{\nu}{2}\right)=-(\nu+1)\frac{d}{d\nu}\left[\ln\Gamma\left(\frac{\nu}{2}\right)-\ln\Gamma\left(\frac{\nu+1}{2}\right)\right]=\frac{\nu+1}{2}\left[\psi\left(\frac{\nu+1}{2}\right)-\psi\left(\frac{\nu}{2}\right)\right]$$
Combining everything together we get the desired result
$$H=\ln \left(\sqrt{\nu}B\left(\frac{1}{2},\frac{\nu}{2}\right)\right)+\frac{\nu+1}{2}\left[\psi\left(\frac{\nu+1}{2}\right)-\psi\left(\frac{\nu}{2}\right)\right]$$
