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Question Number 86247 by M±th+et£s last updated on 27/Mar/20

$${\int }_0^8{\int }_0^{x^{\frac{2}{3}}}\sqrt{x^2+y^2+1}dydx$$

Commented by mathmax by abdo last updated on 27/Mar/20

$$A={\int }_0^8\left({\int }_0^{x^{\frac{2}{3}}}\sqrt{y^2+x^2+1}dy\right)dx={\int }_0^{8}A\left(x\right)dx$$$$A\left(x\right)={\int }_0^{x^{\frac{2}{3}}}\sqrt{y^2+x^2+1}dywedothecyangementy=\sqrt{x^2+1}sh\left(t\right)$$$$\Rightarrow t=argsh\left(\frac{y}{\sqrt{x^2+1}}\right)=ln(\frac{y}{\sqrt{x^2+1}}+\sqrt{1+\frac{y^2}{x^2+1}})$$$$A\left(x\right)={\int }_0^{ln(\frac{x^{\frac{2}{3}}}{\sqrt{x^2+1}}+\sqrt{1+\frac{x^{\frac{4}{3}}}{x^2+1}})}\sqrt{x^2+1}ch\left(t\right)\sqrt{x^2+1}ch\left(t\right)dt$$$$=(x^2+1){\int }_0^{ln\left(\frac{x^{\frac{2}{3}}+\sqrt{x^2+1+x^{\frac{4}{3}}}}{\sqrt{x^2+1}}\right)}\left(\frac{1+ch\left(2t\right)}{2}\right)dt$$$$=\frac{1}{2}(x^2+1)ln\left(\frac{x^{\frac{2}{3}}+\sqrt{x^2+1+x^{\frac{4}{3}}}}{\sqrt{x^2+1}}\right)+\frac{1}{4}(x^2+1){\left[sh\left(2t\right)\right]}_0^{...}$$$$=\frac{1}{2}(x^2+1)ln\left(\frac{x^{\frac{2}{3}}+\sqrt{x^2+1+x^{\frac{4}{3}}}}{\sqrt{x^2+1}}\right)+\frac{1}{4}(x^2+1){\left[\frac{e^{2t}-e^{-2t}}{2}\right]}_0^{...}$$$$becontinued....$$

Commented by M±th+et£s last updated on 27/Mar/20

$$godblessyousir.thanks$$

Commented by mathmax by abdo last updated on 28/Mar/20

$$youarewelcomesir.$$

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