Question Number 73155 by MJS last updated on 06/Nov/19
![reposting a former question... ∫(((x)^(1/5) −1)/( (√x)+1))dx= [t=(x)^(1/(10)) → dx=10(x^9 )^(1/(10)) dx] =10∫((t^9 (t−1))/(t^4 −t^3 +t^2 −t+1))dt= =10∫(t^6 −t^4 −t)dt+10∫((t(t^2 −t+1))/(t^4 −t^3 +t^2 −t+1))dt= =((10)/7)t^7 −2t^5 −5t^2 +(5+(√5))∫(t/(t^2 −((1−(√5))/3)t+1))dt+(5−(√5))∫(t/(t^2 −((1+(√5))/2)t+1))dt= and it′s easy to solve these](https://www.tinkutara.com/question/Q73155.png)
$$\mathrm{reposting}\:\mathrm{a}\:\mathrm{former}\:\mathrm{question}… \\ $$$$\int\frac{\sqrt[{\mathrm{5}}]{{x}}−\mathrm{1}}{\:\sqrt{{x}}+\mathrm{1}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt[{\mathrm{10}}]{{x}}\:\rightarrow\:{dx}=\mathrm{10}\sqrt[{\mathrm{10}}]{{x}^{\mathrm{9}} }{dx}\right] \\ $$$$=\mathrm{10}\int\frac{{t}^{\mathrm{9}} \left({t}−\mathrm{1}\right)}{{t}^{\mathrm{4}} −{t}^{\mathrm{3}} +{t}^{\mathrm{2}} −{t}+\mathrm{1}}{dt}= \\ $$$$=\mathrm{10}\int\left({t}^{\mathrm{6}} −{t}^{\mathrm{4}} −{t}\right){dt}+\mathrm{10}\int\frac{{t}\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)}{{t}^{\mathrm{4}} −{t}^{\mathrm{3}} +{t}^{\mathrm{2}} −{t}+\mathrm{1}}{dt}= \\ $$$$=\frac{\mathrm{10}}{\mathrm{7}}{t}^{\mathrm{7}} −\mathrm{2}{t}^{\mathrm{5}} −\mathrm{5}{t}^{\mathrm{2}} +\left(\mathrm{5}+\sqrt{\mathrm{5}}\right)\int\frac{{t}}{{t}^{\mathrm{2}} −\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{3}}{t}+\mathrm{1}}{dt}+\left(\mathrm{5}−\sqrt{\mathrm{5}}\right)\int\frac{{t}}{{t}^{\mathrm{2}} −\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}{t}+\mathrm{1}}{dt}= \\ $$$$\mathrm{and}\:\mathrm{it}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{these} \\ $$