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Question Number 192470 by Spillover last updated on 19/May/23

Answered by Spillover last updated on 19/May/23

$${\int }_0^{\sqrt{2}}\sqrt{1+{\left(2x\right)}^2}dx$$$$Let2x=\sinh \theta dx=\frac{\cosh \theta d\theta }{2}$$$${\int }_0^{\sqrt{2}}\sqrt{1+{\left(2x\right)}^2}dx{\int }_0^{\sqrt{2}}\sqrt{1+\sinh {}^2\theta }.\frac{\cosh \theta d\theta }{2}dx$$$$\frac{1}{2}\int \cosh {}^2\theta d\theta \frac{1}{2}\int \left(\frac{1+\cosh 2\theta }{2}\right)$$$$\frac{1}{4}\int d\theta +\frac{1}{4}\int \cosh 2\theta =\frac{1}{4}\theta +\frac{1}{8}\sinh 2\theta$$$$\frac{1}{4}\theta +\frac{1}{8}\sinh 2\theta =\frac{1}{4}\theta +\frac{1}{4}\sinh \theta \cosh \theta$$$$2x=\sinh \theta \theta =\sinh {}^{-1}2x\cosh \theta =\sqrt{1+\sinh {}^2\theta }$$$$\frac{1}{4}\sinh {}^{-1}2x+\frac{2x}{4}\sqrt{1+4x^2}$$$$\frac{1}{4}\ln \mid 2x+\sqrt{1+4x^2}\mid +\frac{x}{2}\sqrt{1+4x^2}$$$${[\frac{1}{4}\ln \mid 2x+\sqrt{1+4x^2}\mid +\frac{x}{2}\sqrt{1+4x^2}]}_0^{\sqrt{2}}$$$$\frac{3\sqrt{2}}{2}+\frac{1}{4}\ln \mid 3+2\sqrt{3}\mid -0$$$$\frac{3\sqrt{2}}{2}+\frac{1}{4}\ln \mid 3+2\sqrt{2}\mid$$

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