Question Number 86850 by john santu last updated on 01/Apr/20
$$\int\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} \:\left(\sqrt[{\mathrm{4}\:\:}]{\left(\mathrm{x}^{\mathrm{4}} +\mathrm{1}\right)}\right)^{\mathrm{3}} } \\ $$
Commented by john santu last updated on 01/Apr/20
Answered by MJS last updated on 01/Apr/20
![∫(dx/(x^2 (x^4 +1)^(3/4) ))= [t=(((x^4 +1)^(1/4) )/x) → dx=−x^2 (x^4 +1)^(3/4) dt] =−∫dt=−t=−(((x^4 +1))^(1/4) /x)+C](Q86854.png)
$$\int\frac{{dx}}{{x}^{\mathrm{2}} \left({x}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{3}/\mathrm{4}} }= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\left({x}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{1}/\mathrm{4}} }{{x}}\:\rightarrow\:{dx}=−{x}^{\mathrm{2}} \left({x}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{3}/\mathrm{4}} {dt}\right] \\ $$$$=−\int{dt}=−{t}=−\frac{\sqrt[{\mathrm{4}}]{{x}^{\mathrm{4}} +\mathrm{1}}}{{x}}+{C} \\ $$
Commented by john santu last updated on 01/Apr/20
$$\mathrm{super}\:\mathrm{easy}\:\mathrm{sir}.\:\mathrm{how}\:\mathrm{to}\:\mathrm{know}\: \\ $$$$\mathrm{t}\:=\:\frac{\left(\mathrm{x}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{3}/\mathrm{4}} }{\mathrm{x}}\:?\:\mathrm{by}\:\mathrm{observe}\:? \\ $$
Commented by MJS last updated on 01/Apr/20
![trying... (d/dx)[(x^4 +1)^(1/4) ]=(x^3 /((x^4 +1)^(3/4) )) then knowing that (d/dx)[(u/v)]==((u′v−uv′)/v^2 ) and somehow feeling thar v might be x](Q86858.png)
$$\mathrm{trying}... \\ $$$$\frac{{d}}{{dx}}\left[\left({x}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{1}/\mathrm{4}} \right]=\frac{{x}^{\mathrm{3}} }{\left({x}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{3}/\mathrm{4}} } \\ $$$$\mathrm{then}\:\mathrm{knowing}\:\mathrm{that}\:\frac{{d}}{{dx}}\left[\frac{{u}}{{v}}\right]==\frac{{u}'{v}−{uv}'}{{v}^{\mathrm{2}} } \\ $$$$\mathrm{and}\:\mathrm{somehow}\:\mathrm{feeling}\:\mathrm{thar}\:{v}\:\mathrm{might}\:\mathrm{be}\:{x} \\ $$
Commented by john santu last updated on 01/Apr/20
$$\mathrm{waw}..\mathrm{amazing}\:\mathrm{sir}.\:\mathrm{good}\:\mathrm{feeling} \\ $$