Question: Let $f:[-1,1] \rightarrow \mathbb{R}$ be defined by $$ f(x) = \sqrt{\frac{1-x}{2}}. $$ By multiplying both sides of the equation $$ \frac{1}{\sqrt{1-2xt+t^2}} = \sum_{n=1}^{\infty} P_n(x)t^n $$ by $f(x)$ and integrating both sides with respect to $x,$ show that $$ \frac{1}{2t}\left(1+t-\frac{(1-t)^2}{2 \sqrt {t}}\log \frac{1+\sqrt{t}}{1-\sqrt{t}}\right) = \sum_{n=1}^{\infty}t^n \int_{-1}^{1}f(x)P_n(x)dx. $$
I was not able to prove that. What has been done incorrectly? Below are my steps:
$$\displaystyle \int_{-1}^{1}\dfrac{f(x)}{\sqrt{1-2xt+t^2}}dx$$
$$=\displaystyle\int_{-1}^{1}\dfrac{\sqrt{\dfrac{1-x}{2}}}{\sqrt{1-2xt+t^2}}dx $$
$$=\dfrac{1}{\sqrt{2}} \displaystyle \int_{-1}^{1}\dfrac{1-x}{\sqrt{(1-x)[(1+t^2)-2tx]}}dx$$
$$=-\dfrac{1}{4\sqrt{2}t} \displaystyle \int_{-1}^{1}\dfrac{-(1+t)^2 + 4tx}{\sqrt{(1-x)[(1+t^2)-2tx]}}dx -
\dfrac{1}{4\sqrt{2}t} \displaystyle \int_{-1}^{1}\dfrac{(1+t)^2 + 4t}{\sqrt{(1-x)[(1+t^2)-2tx]}}dx $$
$$=- \dfrac{1}{4\sqrt{2}t}\left[\dfrac{(1-x)[(1+t^2)-2tx]^{\frac{1}{2}}}{\frac{1}{2}}\right]\Big|_{-1}^1- \dfrac{(1-t)^2}{4\sqrt{2}t} \displaystyle \int_{-1}^{1}\dfrac{1}{\sqrt{(1+t^2)-(1+t^2)x+2tx^2}}dx $$
$$= -\dfrac{1}{2\sqrt{2}t} \left[0 - 2{(1 + t^2) + 2t} \right]^{\frac{1}{2}}- \dfrac{(1-t)^2}{4\sqrt{2}t} \displaystyle \int_{-1}^{1} \dfrac{1}{\sqrt{2t}}\dfrac{dx}{\sqrt{\left[x-\dfrac{(1+t)^2}{4t}\right]^2 + \left(\dfrac{1+t^2}{2t}-\dfrac{(1+t)^4}{16t^2}\right)}}$$
$$=\dfrac{(1+t)}{2t}-\dfrac{(1-t)^2}{4\sqrt{2}t}\cdot\dfrac{1}{\sqrt{2t}} \displaystyle \int_{-1}^{1}\dfrac{dx}{\sqrt{\left[x-\dfrac{(1+t)^2}{4t}\right]^2-\dfrac{(1-t)^4}{16t^2}}}$$
$$=\dfrac{(1+t)}{2t} - \dfrac{(1-t)^2}{8t\sqrt{t}} \displaystyle \int_{-1}^{1} \dfrac{dx}{\sqrt{\left[x-\dfrac{(1+t)^2}{4t}\right]^2-\dfrac{(1-t)^4}{16t^2}}}$$
$$=\dfrac{(1+t)}{2t} - \dfrac{(1-t)^2}{8t\sqrt{t}} \cosh^{-1}\left[\dfrac{x- \frac{(1+x)^2}{4t}}{\frac{(1-t)^2}{4t}}\right]\Big|_{-1}^1$$
$$=\dfrac{(1+t)}{2t} - \dfrac{(1-t)^2}{8t\sqrt{t}}\left[\cosh^{-1}(-1)- \cosh^{-1}\left(\dfrac{-1-6t-t^2}{1-2t+t^2}\right)\right]$$
$$=\dfrac{(1+t)}{2t} - \dfrac{(1-t)^2}{8t\sqrt{t}}\ln\left(\dfrac{(1+\sqrt{t})^2}{(1-\sqrt{t})(1+\sqrt{t}}\right)$$
$$=\dfrac{(1+t)}{2t} - \dfrac{(1-t)^2}{8t\sqrt{t}}\ln\left(\dfrac{1+\sqrt{t}}{1-\sqrt{t}}\right)$$
$$= \dfrac{1}{2t}\left(1+t-\dfrac{(1-t)^2}{4 \sqrt {t}}\ln \dfrac{1+\sqrt{t}}{1-\sqrt{t}}\right)$$
