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Question Number 143167 by pticantor last updated on 10/Jun/21

$$\int \mathit{a}\mathit{r}\mathit{c}\mathit{t}\mathit{a}\mathit{n}\left(\sqrt{\sqrt{\mathit{x}}+1}\right)\mathit{d}\mathit{x}=???$$$$$$$${\mathit{p}\mathit{r}\mathit{o}\mathit{p}\mathit{o}\mathit{s}\mathit{e}}^{\prime }\mathit{p}\mathit{a}\mathit{r}\mathit{R}\mathit{o}\mathit{d}\mathit{r}\mathit{i}\mathit{g}\mathit{u}\mathit{e}$$

Answered by Olaf_Thorendsen last updated on 10/Jun/21

$$\mathrm{F}\left(x\right)=\int \arctan \left(\sqrt{\sqrt{x}+1}\right)dx$$$$\mathrm{F}\left(x\right)=x\arctan \left(\sqrt{\sqrt{x}+1}\right)-\int x\frac{\frac{\frac{1}{2\sqrt{x}}}{2\sqrt{\sqrt{x}+1}}}{1+(\sqrt{x}+1)}dx$$$$\mathrm{F}\left(x\right)=x\arctan \left(\sqrt{\sqrt{x}+1}\right)-\frac{1}{4}\int \frac{\sqrt{x}}{(\sqrt{x}+2)\sqrt{\sqrt{x}+1}}dx$$$$\mathrm{G}\left(x\right)=\int \frac{\sqrt{x}}{(\sqrt{x}+2)\sqrt{\sqrt{x}+1}}dx$$$$\mathrm{Let}u=\sqrt{x}$$$$\mathrm{G}\left(u\right)=\int \frac{u}{(u+2)\sqrt{u+1}}2udu$$$$\mathrm{G}\left(u\right)=\int \frac{2u^2}{(u+2)\sqrt{u+1}}du$$$$\mathrm{Let}v=\sqrt{u+1}$$$$\mathrm{G}\left(v\right)=\int \frac{2{(v^2-1)}^2}{(v^2-1+2)v}2vdv$$$$\mathrm{G}\left(v\right)=\int \frac{4{(v^2-1)}^2}{v^2+1}dv$$$$\mathrm{G}\left(v\right)=\int [4v^2-12+\frac{16}{v^2+1}]dv$$$$\mathrm{G}\left(v\right)=\frac{4}{3}{v}^3-12v+16\arctan v$$$$\mathrm{G}\left(u\right)=\frac{4}{3}{(u+1)}^{3/2}-12\sqrt{u+1}+16\arctan \left(\sqrt{u+1}\right)$$$$\mathrm{G}\left(x\right)=\frac{4}{3}{(\sqrt{x}+1)}^{3/2}-12\sqrt{\sqrt{x}+1}+16\arctan \left(\sqrt{\sqrt{x}+1}\right)$$$$\mathrm{F}\left(x\right)=x\arctan \left(\sqrt{\sqrt{x}+1}\right)-\frac{1}{4}\mathrm{G}\left(x\right)$$$$\mathrm{F}\left(x\right)=x\arctan \left(\sqrt{\sqrt{x}+1}\right)-\frac{1}{3}{(\sqrt{x}+1)}^{3/2}$$$$+3\sqrt{\sqrt{x}+1}-4\arctan \left(\sqrt{\sqrt{x}+1}\right)$$$$\mathrm{F}\left(x\right)=(x-4)\arctan \left(\sqrt{\sqrt{x}+1}\right)-\frac{1}{3}{(\sqrt{x}+1)}^{3/2}+3\sqrt{\sqrt{x}+1}$$

Answered by mathmax by abdo last updated on 11/Jun/21

$$\mathrm{\Phi }=\int \arctan \left(\sqrt{\sqrt{\mathrm{x}}+1}\right)\mathrm{dx}\mathrm{changement}\sqrt{\sqrt{\mathrm{x}}+1}=\mathrm{t}\mathrm{give}\sqrt{\mathrm{x}}+1={\mathrm{t}}^2\Rightarrow$$$$\sqrt{\mathrm{x}}={\mathrm{t}}^2-1\Rightarrow \mathrm{x}={({\mathrm{t}}^2-1)}^2\Rightarrow \mathrm{dx}=2\left(2\mathrm{t}\right)({\mathrm{t}}^2-1)=4\mathrm{t}({\mathrm{t}}^2-1)\Rightarrow$$$$\mathrm{\Phi }=\int 4\mathrm{t}({\mathrm{t}}^2-1)\arctan \left(\mathrm{t}\right)\mathrm{dt}=4\int ({\mathrm{t}}^3-\mathrm{t})\arctan \left(\mathrm{t}\right)\mathrm{dt}$$$${=}_{\mathrm{by}\mathrm{parts}}4\{(\frac{{\mathrm{t}}^4}{4}-\frac{{\mathrm{t}}^2}{2})\mathrm{arctant}-\int (\frac{{\mathrm{t}}^4}{4}-\frac{{\mathrm{t}}^2}{2})\frac{\mathrm{dt}}{1+{\mathrm{t}}^2}$$$$=({\mathrm{t}}^4-{2\mathrm{t}}^2)\mathrm{arctant}-\int ({\mathrm{t}}^4-{2\mathrm{t}}^2)\frac{\mathrm{dt}}{1+{\mathrm{t}}^2}\mathrm{but}$$$$\int \frac{{\mathrm{t}}^4-{2\mathrm{t}}^2}{{\mathrm{t}}^2+1}\mathrm{dt}=\int \frac{{\mathrm{t}}^2({\mathrm{t}}^2+1)-{3\mathrm{t}}^2}{{\mathrm{t}}^2+1}\mathrm{dt}$$$$=\int {\mathrm{t}}^2\mathrm{dt}-3\int \frac{{\mathrm{t}}^2+1-1}{{\mathrm{t}}^2+1}\mathrm{dt}=\frac{{\mathrm{t}}^3}{3}-3\mathrm{t}+3\mathrm{arctant}+\mathrm{c}\Rightarrow$$$$\mathrm{\Phi }=({\mathrm{t}}^4-{2\mathrm{t}}^2)\mathrm{arctant}-\frac{{\mathrm{t}}^3}{3}+3\mathrm{t}-3\mathrm{arctant}+\mathrm{C}$$$$=({(\sqrt{\mathrm{x}}+1)}^2-2(\sqrt{\mathrm{x}}+1)-3)\arctan \left(\sqrt{\sqrt{\mathrm{x}}+1}\right)-\frac{1}{3}{\left(\sqrt{\sqrt{\mathrm{x}}+1}\right)}^3+3\left(\sqrt{\sqrt{\mathrm{x}}+1}\right)+\mathrm{C}$$

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