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Question Number 80144 by behi83417@gmail.com last updated on 31/Jan/20

$$\mathbf{solve}\mathbf{for}\mathbf{x}:$$$$\frac{\sqrt{\mathbf{x}}+1}{\sqrt{\mathbf{x}+1}}+{\mathbf{ax}}^2=\mathbf{x}({\mathbf{a}}^2+1)[\mathbf{a}\in \mathbf{R}]$$

Commented by john santu last updated on 31/Jan/20

$$let\sqrt{x+1}=\sec t$$$$x+1=\sec {}^{2}t\Rightarrow x=\tan {}^{2}t$$$$\frac{\tan t+1}{\sec t}+a\left(\tan {}^{4}t\right)=(a^2+1)\tan {}^{2}t$$$$\cos t(\tan t+1)+a\tan {}^{4}t=(a^2+1)\tan {}^{2}t$$

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