Evaluate the Integral: $\displaystyle\int \sqrt{2x+\sqrt{2x+\sqrt{2x+...\infty}}}dx$, where $\sqrt{2x+\sqrt{2x+\sqrt{2x+...\infty}}}$ is a convergent infinitely nested radical.
2026-05-14 17:48:58.1778780938
Evaluate: $\int \sqrt{2x+\sqrt{2x+\sqrt{2x+...\infty}}}dx$
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$$I=\displaystyle\int \sqrt{2x+\sqrt{2x+\sqrt{2x+...\infty}}} dx$$ Let, $\sqrt{2x+\sqrt{2x+\sqrt{2x+...\infty}}}=u\implies \sqrt{2x+u}=u\implies dx=\dfrac{2u-1}{2}du$ $$\implies I=\displaystyle \int u\cdot \bigg(\dfrac{2u-1}{2}\bigg)du$$ $$\implies I=\displaystyle \int u^{2}-\dfrac{u}{2}du$$ $$\implies I=\displaystyle \int u^{2}-\int \dfrac{u}{2}du$$ $$\implies I=\dfrac{u^{3}}{3}-\dfrac{u^{2}}{4}+C$$