Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 79612 by john santu last updated on 26/Jan/20

∫ (dx/((√(x ))((x)^(1/(4 )) +1)^(10) )) = ?

$$\int\:\frac{\mathrm{dx}}{\sqrt{\mathrm{x}\:}\left(\sqrt[{\mathrm{4}\:}]{\mathrm{x}}+\mathrm{1}\right)^{\mathrm{10}} }\:=\:? \\ $$

Answered by MJS last updated on 26/Jan/20

∫(dx/(x^(1/2) (x^(1/4) +1)^(10) ))=       [t=x^(1/4)  → dx=4x^(3/4) dt]  =4∫(t/((t+1)^(10) ))dt=4∫(dt/((t+1)^9 ))−4∫(dt/((t+1)^(10) ))=  =−(1/(2(t+1)^8 ))+(4/(9(t+1)^9 ))=−((9t+1)/(18(t+1)^9 ))=  =−((9x^(1/4) +1)/(18(x^(1/4) +1)^9 ))+C

$$\int\frac{{dx}}{{x}^{\frac{\mathrm{1}}{\mathrm{2}}} \left({x}^{\frac{\mathrm{1}}{\mathrm{4}}} +\mathrm{1}\right)^{\mathrm{10}} }= \\ $$$$\:\:\:\:\:\left[{t}={x}^{\frac{\mathrm{1}}{\mathrm{4}}} \:\rightarrow\:{dx}=\mathrm{4}{x}^{\frac{\mathrm{3}}{\mathrm{4}}} {dt}\right] \\ $$$$=\mathrm{4}\int\frac{{t}}{\left({t}+\mathrm{1}\right)^{\mathrm{10}} }{dt}=\mathrm{4}\int\frac{{dt}}{\left({t}+\mathrm{1}\right)^{\mathrm{9}} }−\mathrm{4}\int\frac{{dt}}{\left({t}+\mathrm{1}\right)^{\mathrm{10}} }= \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}\left({t}+\mathrm{1}\right)^{\mathrm{8}} }+\frac{\mathrm{4}}{\mathrm{9}\left({t}+\mathrm{1}\right)^{\mathrm{9}} }=−\frac{\mathrm{9}{t}+\mathrm{1}}{\mathrm{18}\left({t}+\mathrm{1}\right)^{\mathrm{9}} }= \\ $$$$=−\frac{\mathrm{9}{x}^{\frac{\mathrm{1}}{\mathrm{4}}} +\mathrm{1}}{\mathrm{18}\left({x}^{\frac{\mathrm{1}}{\mathrm{4}}} +\mathrm{1}\right)^{\mathrm{9}} }+{C} \\ $$

Commented by peter frank last updated on 26/Jan/20

thanks

$${thanks} \\ $$

Commented by john santu last updated on 27/Jan/20

thank you mister

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{mister} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com