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 166723 by cortano1 last updated on 26/Feb/22

$$\int \frac{\sqrt[5]{\mathrm{x}}-1}{\sqrt{\mathrm{x}}+1}\mathrm{dx}=?$$

Answered by MJS_new last updated on 26/Feb/22

$$\int \frac{x^{1/5}-1}{x^{1/2}+1}dx=$$$$[t=x^{1/10}\to dx=10t^{9}dt]$$$$=10\int \frac{t^9(t^2-1)}{t^5+1}dt=10\int \frac{t^9(t-1)}{t^4-t^3+t^2-t+1}dt=$$$$=10\int \frac{t^9(t-1)}{(t^2-\frac{1+\sqrt{5}}{2}t+1)(t^2-\frac{1-\sqrt{5}}{2}t+1)}dt=$$$$=I_1+I_2+I_3+I_4+I_5$$$$\mathrm{with}$$$$I_1=10\int (t^6-t^4-t)dt=\frac{10}{7}{t}^7-2t^5-5t^2$$$$I_2=\int \frac{(5-\sqrt{5})t-\sqrt{5}}{t^2-\frac{1+\sqrt{5}}{2}t+1}dt=\frac{5-\sqrt{5}}{2}\ln (t^2-\frac{1+\sqrt{5}}{2}t+1)$$$$I_3=\int \frac{(5+\sqrt{5})t+\sqrt{5}}{t^2-\frac{1-\sqrt{5}}{2}t+1}dt=\frac{5+\sqrt{5}}{2}\ln (t^2-\frac{1-\sqrt{5}}{2}t+1)$$$$I_4=\sqrt{5}\int \frac{dt}{t^2-\frac{1+\sqrt{5}}{2}t+1}=\sqrt{2(5+\sqrt{5})}\arctan \frac{\sqrt{10(5+\sqrt{5})}(t-\frac{1+\sqrt{5}}{4})}{5}$$$$I_5=-\sqrt{5}\int \frac{dt}{t^2-\frac{1-\sqrt{5}}{2}t+1}=-\sqrt{2(5-\sqrt{5})}\arctan \frac{\sqrt{10(5-\sqrt{5})}(t-\frac{1-\sqrt{5}}{4})}{5}$$$$...$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com