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Question Number 199385 by hardmath last updated on 02/Nov/23

$$\mathrm{Find}:$$$$\mathrm{\Omega }={\int }_0^1{\mathrm{x}}^{15}\sqrt{1+{3\mathrm{x}}^8}\mathrm{dx}=?$$

Answered by witcher3 last updated on 02/Nov/23

$${\mathrm{x}}^{15}={\mathrm{x}}^{7.}{\mathrm{x}}^8=\frac{1}{8}{\mathrm{x}}^8\mathrm{d}\left({\mathrm{x}}^8\right)$$$$\int \mathrm{u}\sqrt{1+\mathrm{u}}=\int {(1+\mathrm{u})}^{\frac{3}{2}}-{(1+\mathrm{u})}^{\frac{1}{2}}\mathrm{du}....$$

Commented by hardmath last updated on 02/Nov/23

$$\mathrm{thank}\mathrm{you}\mathrm{ser},\mathrm{but}\mathrm{how}$$

Commented by witcher3 last updated on 02/Nov/23

$$\mathrm{u}={\mathrm{x}}^8$$$$\iff \frac{1}{8}{\int }_0^1\mathrm{u}\sqrt{1+3\mathrm{u}}.\mathrm{du}$$$$=\frac{1}{24}{\int }_0^1(3\mathrm{u}+1-1){(1+3\mathrm{u})}^{\frac{1}{2}}\mathrm{du}$$$$=\frac{1}{24}\{{\int }_0^1{(3\mathrm{u}+1)}^{\frac{3}{2}}\mathrm{du}-{\int }_0^1{(1+3\mathrm{u})}^{\frac{1}{2}}\mathrm{du}$$$$=\frac{1}{24}{[\frac{1}{3(\frac{3}{2}+1)}{(3\mathrm{u}+1)}^{\frac{5}{2}}-\frac{1}{3\left(\frac{3}{2}\right)}{(1+3\mathrm{u})}^{\frac{3}{2}}]}_0^1$$$$=\frac{1}{24}[\frac{2}{15}{\left(2\right)}^5-\frac{2}{15}-\frac{2}{9}{2}^3+\frac{2}{9}]$$

Commented by York12 last updated on 03/Nov/23

$$\mathrm{Where}\mathrm{are}\mathrm{you}\mathrm{from}$$

Commented by witcher3 last updated on 03/Nov/23

$$\mathrm{i}\mathrm{m}\mathrm{from}\mathrm{france}$$

Commented by York12 last updated on 03/Nov/23

$$\mathrm{ur}\mathrm{speaking}\mathrm{arabic}\mathrm{lol}$$

Commented by witcher3 last updated on 03/Nov/23

$$\mathrm{yes}\mathrm{i}\mathrm{speak}\mathrm{arabic}\mathrm{im}\mathrm{algerian}\mathrm{livin}\mathrm{in}\mathrm{France}$$

Commented by York12 last updated on 03/Nov/23

$$\mathrm{hmmmm}\mathrm{okay}\mathrm{nice}\mathrm{to}\mathrm{meet}\mathrm{you}:),\mathrm{I}\mathrm{am}\mathrm{muslim}$$$$\mathrm{too}.$$

Commented by York12 last updated on 03/Nov/23

$$\mathrm{wow}\mathrm{ur}\mathrm{algerian},\mathrm{nice}\mathrm{to}\mathrm{hear}\mathrm{that},\mathrm{I}\mathrm{am}\mathrm{saudi}$$

Commented by Frix last updated on 03/Nov/23

$$\mathrm{I}\mathrm{once}\mathrm{had}\mathrm{a}\mathrm{student}\mathrm{who}\mathrm{stated}‘‘\mathrm{I}\mathrm{learn}$$$$\mathrm{Pythagoras}\mathrm{just}\mathrm{for}\mathrm{school}\mathrm{because}\mathrm{I}\mathrm{have}$$$$\mathrm{to}.\mathrm{All}\mathrm{pagans}\mathrm{and}\mathrm{all}\mathrm{their}\mathrm{friends}\mathrm{go}\mathrm{to}$$$$\mathrm{hell}{.}^{″}$$

Commented by York12 last updated on 03/Nov/23

$$\mathrm{LMAO}$$

Commented by witcher3 last updated on 04/Nov/23

$$\mathrm{nice}\mathrm{To}\mathrm{meet}\mathrm{You}\mathrm{yorke}$$$$\mathrm{selam}\mathrm{alaykoum}$$

Commented by witcher3 last updated on 04/Nov/23

$$\mathrm{lol}\mathrm{frix}$$

Commented by York12 last updated on 04/Nov/23

$$\mathrm{alaykom}\mathrm{alsalam}\mathrm{nice}\mathrm{to}\mathrm{meet}\mathrm{you}\mathrm{too}$$$$\mathrm{my}\mathrm{name}\mathrm{is}\mathrm{YUSUF}\mathrm{btw}$$

Answered by Frix last updated on 03/Nov/23

$$\int x^{15}\sqrt{3x^8+1}dx\stackrel{t=3x^8+1}{=}$$$$=\frac{1}{72}\int (t^{\frac{3}{2}}-t^{\frac{1}{2}})dt=$$$$=\frac{t^{\frac{5}{2}}}{180}-\frac{t^{\frac{3}{2}}}{108}=\frac{t^{\frac{3}{2}}(3t-5)}{540}=$$$$=\frac{{(3x^8+1)}^{\frac{3}{2}}(9x^8-2)}{540}$$$$\mathrm{\Omega }=\frac{29}{270}$$

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