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Question Number 66627 by behi83417@gmail.com last updated on 17/Aug/19

Commented by behi83417@gmail.com last updated on 17/Aug/19

$$\text{You can't use 'macro parameter character \#' in math mode}$$

Commented by mathmax by abdo last updated on 18/Aug/19

$$1)letI=\int \sqrt{1+x+x^2+x^3}dx\Rightarrow I=\int \sqrt{1+x+x^2(1+x)}dx$$$$=\int \sqrt{(1+x)(1+x^2)}dx=\int \sqrt{1+x}\sqrt{1+x^2}dxchangement\sqrt{1+x}=t$$$$give1+x=t^2\Rightarrow x=t^2-1\Rightarrow I=\int t\sqrt{1+{(t^2-1)}^2}\left(2t\right)dt$$$$=2\int t^2\sqrt{1+{(t^2-1)}^2}dtlet{t}^2-1=sh\left(u\right)\Rightarrow t^2=1+sh\left(u\right)\Rightarrow$$$$2tdt=chdu\Rightarrow I=\int \sqrt{1+sh\left(u\right)}ch\left(u\right)ch\left(u\right)du$$$$=\int {ch}^2\left(u\right)\sqrt{1+sh\left(u\right)}du=\int ch\left(u\right)ch\left(u\right)(1+sh{\left(u\right)}^{\frac{1}{2}}du$$$$bypartsf=chuand{g}^{{}^{\prime }}=ch\left(u\right){(1+sh\left(u\right))}^{\frac{1}{2}}\Rightarrow g=\frac{2}{3}{(1+sh\left(u\right))}^{\frac{3}{2}}$$$$\Rightarrow I=\frac{2}{3}ch\left(u\right){(1+sh\left(u\right))}^{\frac{3}{2}}-\frac{2}{3}\int sh\left(u\right){(1+sh\left(u\right))}^{\frac{3}{2}}du$$$$....becontinued...$$

Commented by Tanmay chaudhury last updated on 18/Aug/19

$$challengingintregal...$$

Commented by Tanmay chaudhury last updated on 18/Aug/19

$$sirBehiplsuploadthesolution...ifavailable$$$$theseproblemssnatchedoursleep..$$

Answered by mind is power last updated on 18/Aug/19

$$thefirst{can}^{\prime }tbeexpressedusualfunctionelepticintegral$$$$$$

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