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Question Number 67471 by AnjanDey last updated on 27/Aug/19

$$\mathit{E}\mathit{v}\mathit{a}\mathit{l}\mathit{u}\mathit{a}\mathit{t}\mathit{e}:\int \sqrt{x\sqrt{x+1}}dx$$

Commented by MJS last updated on 28/Aug/19

$$\int \sqrt{x\sqrt{x+1}}dx=$$$$[t=\sqrt{x}\to dx=2\sqrt{x}dt]$$$$=2\int t^2{(t^2+1)}^{\frac{1}{4}}dt=$$$$=-\frac{4}{21}\int \frac{dt}{{(t^2+1)}^{\frac{3}{4}}}+2\int \frac{t^4+t^2+\frac{2}{21}}{{(t^2+1)}^{\frac{3}{4}}}dt$$$$=-\frac{4}{21}\int \frac{dt}{{(t^2+1)}^{\frac{3}{4}}}+\frac{4}{21}t(3t^2+1){(t^2+1)}^{\frac{1}{4}}$$$$...\mathrm{I}\mathrm{cannot}\mathrm{solve}\mathrm{the}{1}^{\mathrm{st}}\mathrm{one}$$$$\mathrm{plus}\mathrm{I}\mathrm{somehow}\mathrm{solved}\mathrm{the}{2}^{\mathrm{nd}}\mathrm{one}\mathrm{but}\mathrm{I}\mathrm{really}$$$${\mathrm{don}}^{\prime }\mathrm{t}\mathrm{remember}\mathrm{how}...\mathrm{I}\mathrm{lost}\mathrm{some}\mathrm{papers}...$$$$\mathrm{anyway}\frac{d}{dt}\left[\frac{4}{21}t(3t^2+1){(t^2+1)}^{\frac{1}{4}}\right]=2\frac{t^4+t^2+\frac{2}{21}}{{(t^2+1)}^{\frac{3}{4}}}$$$$\mathrm{and}-\frac{4}{21{(t^2+1)}^{\frac{3}{4}}}+2\frac{t^4+t^2+\frac{2}{21}}{{(t^2+1)}^{\frac{3}{4}}}=2t^2{(t^2+1)}^{\frac{1}{4}}$$

Commented by Abdo msup. last updated on 28/Aug/19

$$lettakeatrywedothechangement\sqrt{x+1}=t\Rightarrow$$$$x+1=t^2\Rightarrow x=t^2-1\Rightarrow$$$$\int \sqrt{x\sqrt{x+1}}dx=\int \sqrt{(t^2-1)t}\left(2t\right)dt$$$$=2\int t\sqrt{t^3-t}dt=2\int t\sqrt{t(t^2-1)}dt$$$$=2\int t^{\frac{3}{2}}\sqrt{t^2-1}dtandchangementt=ch\left(u\right)give$$$$I=2\int {\left(chu\right)}^{\frac{3}{2}}sh\left(u\right)sh\left(u\right)du$$$$bypsrts{f}^{{}^{\prime }}=sh\left(u\right){\left(chu\right)}^{\frac{3}{2}}andg=shu$$$$I=\frac{2}{5}{\left(chu\right)}^{\frac{5}{2}}sh\left(u\right)-\int \frac{2}{5}{\left(chu\right)}^{\frac{5}{2}}ch\left(u\right)du$$$$=\frac{2}{5}{\left(chu\right)}^{\frac{5}{2}}-\frac{2}{5}\int {\left(chu\right)}^{\frac{7}{2}}du$$$$theproblemeishowtocalculateintehralat$$$$form\int {\left(chu\right)}^{r}duwithr\in Q^{+}?...becontinued...$$

Commented by Abdo msup. last updated on 28/Aug/19

$$sirmjsthisintegralisnotsimpletosolve...$$

Commented by MJS last updated on 28/Aug/19

$$\mathrm{thank}\mathrm{you}\mathrm{for}\mathrm{trying}$$

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