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Question Number 210082 by klipto last updated on 30/Jul/24

$$\mathbf{prove}\mathbf{that}$$$${\int }_0^1\frac{\mathbf{ln}(1-\mathbf{t}+{\mathbf{tx}}^2)}{{\mathbf{x}}^2-1}\mathrm{dx}={\left({\mathbf{sin}}^{-1}\left(\sqrt{\mathbf{t}}\right)\right)}^2$$$$\mathbf{please}\mathbf{anyone}..$$

Answered by Frix last updated on 30/Jul/24

$$\mathrm{Simply}\mathrm{use}{\mathrm{Feynman}}^{\prime }\mathrm{s}\mathrm{Technique}$$$$\underset{0}{\stackrel{1}{\int }}\frac{\ln ({tx}^2-t+1)}{x^2-1}dx=f\left(t\right)$$$$f\left(0\right)=0$$$$f^{\prime }\left(t\right)=\underset{0}{\stackrel{1}{\int }}\frac{\partial }{\partial t}\left[\frac{\ln ({tx}^2-t+1)}{x^2-1}\right]dx=$$$$=\underset{0}{\stackrel{1}{\int }}\frac{dx}{{tx}^2-t+1}\stackrel{u=\frac{\sqrt{t}}{\sqrt{1-t}}x}{=}$$$$=\frac{1}{\sqrt{t}\sqrt{1-t}}\underset{0}{\stackrel{\frac{\sqrt{t}}{\sqrt{1-t}}}{\int }}\frac{du}{u^2+1}=$$$$=\frac{1}{\sqrt{t}\sqrt{1-t}}{\left[{\tan }^{-1}u\right]}_0^{\frac{\sqrt{t}}{\sqrt{1-t}}}=\frac{{\tan }^{-1}\frac{\sqrt{t}}{\sqrt{1-t}}}{\sqrt{t}\sqrt{1-t}}=$$$$=\frac{{\sin }^{-1}\sqrt{t}}{\sqrt{t}\sqrt{1-t}}$$$$f\left(t\right)=\int \frac{{\sin }^{-1}\sqrt{t}}{\sqrt{t}\sqrt{1-\alpha t}}dt\stackrel{v={\sin }^{-1}\sqrt{t}}{=}2\int vdv=v^2=$$$$={\left({\sin }^{-1}\sqrt{t}\right)}^2+C$$$$f\left(0\right)=0\Rightarrow C=0$$$$f\left(t\right)={\left({\sin }^{-1}\sqrt{t}\right)}^2$$

Commented by klipto last updated on 30/Jul/24

$$\mathbf{thanks}\mathbf{my}\mathbf{bro}$$

Commented by klipto last updated on 30/Jul/24

$$\mathbf{which}\mathbf{pdf}\mathbf{do}\mathbf{you}\mathbf{have}\mathbf{on}\mathbf{this}\mathbf{or}\mathbf{channel}$$

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