Question Number 79612 by john santu last updated on 26/Jan/20
$$\int\:\frac{\mathrm{dx}}{\:\sqrt{\mathrm{x}\:}\left(\sqrt[{\mathrm{4}\:}]{\mathrm{x}}+\mathrm{1}\right)^{\mathrm{10}} }\:=\:? \\ $$
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](https://www.tinkutara.com/question/Q79622.png)
$$\int\frac{{dx}}{{x}^{\frac{\mathrm{1}}{\mathrm{2}}} \left({x}^{\frac{\mathrm{1}}{\mathrm{4}}} +\mathrm{1}\right)^{\mathrm{10}} }= \\ $$$$\:\:\:\:\:\left[{t}={x}^{\frac{\mathrm{1}}{\mathrm{4}}} \:\rightarrow\:{dx}=\mathrm{4}{x}^{\frac{\mathrm{3}}{\mathrm{4}}} {dt}\right] \\ $$$$=\mathrm{4}\int\frac{{t}}{\left({t}+\mathrm{1}\right)^{\mathrm{10}} }{dt}=\mathrm{4}\int\frac{{dt}}{\left({t}+\mathrm{1}\right)^{\mathrm{9}} }−\mathrm{4}\int\frac{{dt}}{\left({t}+\mathrm{1}\right)^{\mathrm{10}} }= \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}\left({t}+\mathrm{1}\right)^{\mathrm{8}} }+\frac{\mathrm{4}}{\mathrm{9}\left({t}+\mathrm{1}\right)^{\mathrm{9}} }=−\frac{\mathrm{9}{t}+\mathrm{1}}{\mathrm{18}\left({t}+\mathrm{1}\right)^{\mathrm{9}} }= \\ $$$$=−\frac{\mathrm{9}{x}^{\frac{\mathrm{1}}{\mathrm{4}}} +\mathrm{1}}{\mathrm{18}\left({x}^{\frac{\mathrm{1}}{\mathrm{4}}} +\mathrm{1}\right)^{\mathrm{9}} }+{C} \\ $$
Commented by peter frank last updated on 26/Jan/20
$${thanks} \\ $$
Commented by john santu last updated on 27/Jan/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{mister} \\ $$