Question Number 188557 by MathsFan last updated on 03/Mar/23
$$\int\sqrt{\boldsymbol{{x}}\sqrt{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}}}\boldsymbol{{dx}} \\ $$
Answered by MJS_new last updated on 03/Mar/23
![∫x^(1/2) (x^2 +1)^(1/4) dx= [t=(x^(1/2) /((x^2 +1)^(1/4) )) → dx=2x^(1/2) (x^2 +1)^(5/4) dt] =2∫(t^2 /((t^4 −1)^2 ))dt now decompose etc. I get: −(t^3 /(2(t^4 −1)))+(1/8)ln ((t+1)/(t−1)) +(1/4)arctan t the rest is easy: (1/2)x^(3/2) (x^2 +1)^(1/4) +(1/8)ln (x^(1/2) +(x^2 +1)^(1/4) ) −(1/(16))ln x +(1/4)arctan (((x^2 +1)^(1/4) )/x^(1/2) ) +C](https://www.tinkutara.com/question/Q188594.png)
$$\int{x}^{\mathrm{1}/\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{1}/\mathrm{4}} {dx}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{{x}^{\mathrm{1}/\mathrm{2}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{1}/\mathrm{4}} }\:\rightarrow\:{dx}=\mathrm{2}{x}^{\mathrm{1}/\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{5}/\mathrm{4}} {dt}\right] \\ $$$$=\mathrm{2}\int\frac{{t}^{\mathrm{2}} }{\left({t}^{\mathrm{4}} −\mathrm{1}\right)^{\mathrm{2}} }{dt} \\ $$$$\mathrm{now}\:\mathrm{decompose}\:\mathrm{etc}.\:\mathrm{I}\:\mathrm{get}: \\ $$$$−\frac{{t}^{\mathrm{3}} }{\mathrm{2}\left({t}^{\mathrm{4}} −\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{ln}\:\frac{{t}+\mathrm{1}}{{t}−\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{arctan}\:{t} \\ $$$$\mathrm{the}\:\mathrm{rest}\:\mathrm{is}\:\mathrm{easy}: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{3}/\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{1}/\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{8}}\mathrm{ln}\:\left({x}^{\mathrm{1}/\mathrm{2}} +\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{1}/\mathrm{4}} \right)\:−\frac{\mathrm{1}}{\mathrm{16}}\mathrm{ln}\:{x}\:+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{arctan}\:\frac{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{1}/\mathrm{4}} }{{x}^{\mathrm{1}/\mathrm{2}} }\:+{C} \\ $$