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Question Number 46101 by Saorey last updated on 21/Oct/18

$$\mathrm{I}=\int \frac{{\mathrm{x}}^{\mathrm{n}+1}}{\sqrt{1+{\mathrm{x}}^{\mathrm{n}}}}\mathrm{dx}=?$$

Answered by tanmay.chaudhury50@gmail.com last updated on 21/Oct/18

$$x^n={tan}^2\theta so{nx}^{n-1}dx=2tan\theta {sec}^2\theta d\theta$$$$dx=\frac{2}{n}\times \frac{tan\theta {sec}^2\theta }{{\left[{\left(tan\theta \right)}^{\frac{2}{n}}\right]}^{n-1}}d\theta$$$$\int \frac{{\left[{\left(tan\theta \right)}^{\frac{2}{n}}\right]}^{n+1}}{sec\theta }\times \frac{2}{n}\times \frac{tan\theta {sec}^2\theta }{{\left[{\left(tan\theta \right)}^{\frac{2}{n}}\right]}^{n-1}}d\theta$$$$\frac{2}{n}\int tan\theta sec\theta \times {\left(tan\theta \right)}^{\frac{2n+2}{n}-\frac{2n-2}{n}}d\theta$$$$\frac{2}{n}\int tan\theta sec\theta {tan}^{\frac{4}{n}}\theta d\theta$$$$\frac{2}{n}\int tan\theta sec\theta {\left({tan}^2\theta \right)}^{\frac{2}{n}}d\theta$$$$\frac{2}{n}\int tan\theta sec\theta {({sec}^2\theta -1)}^{\frac{2}{n}}d\theta$$$$k=sec\theta dk=sec\theta tan\theta d\theta$$$$\frac{2}{n}\int {(k^2-1)}^{\frac{2}{n}}dk$$$$p=\frac{2}{n}$$$$p\int {(k^2-1)}^{p}dk={pI}_p$$$$I_p=\int {(k^2-1)}^{p}dk$$$$={(k^2-1)}^{p}k-\int p{(k^2-1)}^{p-1}\times 2k\times kdk$$$$={(k^2-1)}^{p}k-2p\int {(k^2-1)}^{p-1}(k^2-1+1)dk$$$$=k{(k^2-1)}^p-2p[\int {(k^2-1)}^p+\int {(k^2-1)}^{p-1}]dk$$$$=k{(k^2-1)}^p-2p\times I_p-2{pI}_{p-1}$$$$(2p+1)I_p=k{(k^2-1)}^p-2{pI}_{p-1}$$$$I_p=\frac{k}{2p+1}{(k^2-1)}^p-\frac{2p}{2p+1}{I}_{p-1}$$$${pI}_p=\frac{kp}{2p+1}{(k^2-1)}^p-\frac{2p^2}{2p+2}{I}_{p-1}$$$$\frac{2}{n}{I}_{\frac{2}{n}}=\frac{k\left(\frac{2}{n}\right)}{2\left(\frac{2}{n}\right)+1}{(k^2-1)}^{\frac{2}{n}}-\frac{2{\left(\frac{2}{n}\right)}^2}{2\left(\frac{2}{n}\right)+2}{I}_{\frac{2}{n}-1}$$$$\frac{2}{n}{I}_{\frac{2}{n}}=\frac{\frac{2}{n}}{\frac{4}{n}+1}sec\theta \times {tan}^{\frac{4}{n}}\theta -\frac{\frac{8}{n^2}}{\frac{4}{n}+2}{I}_{\frac{2}{n}-1}$$$$contd...$$$$$$

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