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 43991 by Tawa1 last updated on 19/Sep/18

$$\int \frac{1}{{\mathrm{x}}^{\frac{1}{2}}+{\mathrm{x}}^{\frac{1}{3}}}\mathrm{dx}$$

Answered by Joel578 last updated on 19/Sep/18

$$u=x^{\frac{1}{6}}\to u^6=x$$$$6u^{5}du=dx$$$$$$$$I=\int \frac{6u^5}{u^3+u^2}du$$$$=\int (6u^2-6u-\frac{6}{u+1}+6)du$$

Answered by MJS last updated on 19/Sep/18

$$\int \frac{dx}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}=\int \frac{dx}{x^{\frac{2}{6}}(x^{\frac{1}{6}}+1)}=$$$$[t=x^{\frac{1}{6}}\to dx=6x^{\frac{5}{6}}dt]$$$$=6\int \frac{t^3}{(t+1)}dt=$$$$[u=t+1\to du=dt]$$$$=6\int \frac{{(u-1)}^3}{u}du=6\int (u^2-3u+3-\frac{1}{u})du=$$$$=6(\frac{u^3}{3}-\frac{3u^2}{2}+3u-\ln u)=2u^3-9u^2+18u-6\ln u=$$$$=2{(t+1)}^3-9{(t+1)}^2+18(t+1)-6\ln (t+1)=$$$$=2t^3-3t^2+6t+11-6\ln (t+1)=$$$$=2x^{\frac{1}{2}}-3x^{\frac{1}{3}}+6x^{\frac{1}{6}}+11-6\ln (x^{\frac{1}{6}}+1)=$$$$\left[11\mathrm{as}\mathrm{a}\mathrm{constant}\mathrm{merges}\mathrm{in}C\right]$$$$=2x^{\frac{1}{2}}-3x^{\frac{1}{3}}+6x^{\frac{1}{6}}-6\ln (x^{\frac{1}{6}}+1)+C$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com