Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 191624 by Shrinava last updated on 27/Apr/23

Answered by ARUNG_Brandon_MBU last updated on 28/Apr/23

$${\int }_1^x\sqrt{\frac{t}{1+t^3}}dt$$$$={\int }_1^x\frac{t^{\frac{1}{2}}}{\sqrt{1+{\left(t^{\frac{3}{2}}\right)}^2}}dt=\frac{2}{3}{\int }_1^x\frac{d\left(t^{\frac{3}{2}}\right)}{\sqrt{1+{\left(t^{\frac{3}{2}}\right)}^2}}$$$$=\frac{2}{3}{\left[\mathrm{argsh}\left(t^{\frac{3}{2}}\right)\right]}_1^x={[\frac{2}{3}\ln \mid \sqrt{t^3}+\sqrt{1+t^3}\mid ]}_1^x$$$$=\frac{2}{3}\ln (\sqrt{x^3}+\sqrt{1+x^3})-\frac{2}{3}\ln (1+\sqrt{2})$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com