(1)

\begin{align*} H_{\frac{1}{2}} & =\int_{0}^{1}\frac{1-x^{\frac{1}{2}}}{1-x}dx\\ & =2\int_{0}^{1}\frac{1-t}{1-t^{2}}tdt\\ & =2\int_{0}^{1}\frac{t}{1+t}dt\\ & =2\int_{0}^{1}\left(1-\frac{1}{1+t}\right)dt\\ & =2\left[t-\log\left(1+t\right)\right]_{0}^{1}\\ & =2-2\log2 \end{align*}

(2)

\begin{align*} H_{\frac{1}{3}} & =\int_{0}^{1}\frac{1-x^{\frac{1}{3}}}{1-x}dx\\ & =3\int_{0}^{1}\frac{1-t}{1-t^{3}}t^{2}dt\\ & =3\int_{0}^{1}\frac{t^{2}}{t^{2}+t+1}dt\\ & =3\int_{0}^{1}\left(1-\frac{t+1}{t^{2}+t+1}\right)dt\\ & =3\int_{0}^{1}\left(1-\frac{1}{2}\frac{2t+1}{t^{2}+t+1}-\frac{1}{2}\frac{1}{t^{2}+t+1}\right)dt\\ & =3\int_{0}^{1}\left(1-\frac{1}{2}\frac{2t+1}{t^{2}+t+1}-\frac{1}{2}\frac{1}{\frac{3}{4}+\left(t+\frac{1}{2}\right)^{2}}\right)dt\\ & =3\left[t-\frac{1}{2}\log\left(t^{2}+t+1\right)-\frac{1}{2}\frac{2}{\sqrt{3}}\tan^{\bullet}\frac{2\left(t+\frac{1}{2}\right)}{\sqrt{3}}\right]_{0}^{1}\\ & =3-\frac{3}{2}\log3-\frac{\sqrt{3}\pi}{6} \end{align*}

(3)

\begin{align*} H_{\frac{2}{3}} & =H_{\frac{1}{3}}+\pi\tan^{-1}\frac{\pi}{3}-3+\frac{1}{1-\frac{1}{3}}\cmt{\because H_{1-z}-H_{z}=\pi\tan^{-1}\left(\pi z\right)+\frac{1}{1-z}-\frac{1}{z}}\\ & =3-\frac{3}{2}\log3-\frac{\sqrt{3}\pi}{6}+\frac{\pi}{\sqrt{3}}-3+\frac{3}{2}\\ & =\frac{3}{2}+\frac{3}{2}\log3+\frac{\sqrt{3}\pi}{6} \end{align*}

(4)

\begin{align*} H_{\frac{1}{4}} & =\int_{0}^{1}\frac{1-x^{\frac{1}{4}}}{1-x}dx\\ & =4\int_{0}^{1}\frac{1-t}{1-t^{4}}t^{3}dt\\ & =4\int_{0}^{1}\frac{t^{3}}{t^{3}+t^{2}+t+1}dt\\ & =4\int_{0}^{1}\left(1-\frac{t^{2}+t+1}{t^{3}+t^{2}+t+1}\right)dt\\ & =4\int_{0}^{1}\left(1-\frac{t+1}{2\left(t^{2}+1\right)}-\frac{1}{2\left(t+1\right)}\right)dt\\ & =4\int_{0}^{1}\left(1-\frac{2t}{4\left(t^{2}+1\right)}-\frac{1}{2\left(t^{2}+1\right)}-\frac{1}{2\left(t+1\right)}\right)dt\\ & =4\left[t-\frac{1}{4}\log\left(t^{2}+1\right)-\frac{1}{2}\tan^{\bullet}t-\frac{1}{2}\log\left(t+1\right)\right]_{0}^{1}\\ & =4-\frac{\pi}{2}-3\log2 \end{align*}

(5)

\begin{align*} H_{\frac{3}{4}} & =H_{\frac{1}{4}}+\pi\tan^{-1}\frac{\pi}{4}-4+\frac{1}{1-\frac{1}{4}}\cmt{\because H_{1-z}-H_{z}=\pi\tan^{-1}\left(\pi z\right)+\frac{1}{1-z}-\frac{1}{z}}\\ & =4-\frac{\pi}{2}-3\log2+\pi-4+\frac{4}{3}\\ & =\frac{4}{3}+\frac{\pi}{2}-3\log2 \end{align*}