Question Number 191624 by Shrinava last updated on 27/Apr/23
Answered by ARUNG_Brandon_MBU last updated on 28/Apr/23
![∫_1 ^x (√(t/(1+t^3 )))dt =∫_1 ^x (t^(1/2) /( (√(1+(t^(3/2) )^2 ))))dt=(2/3)∫_1 ^x ((d(t^(3/2) ))/( (√(1+(t^(3/2) )^2 )))) =(2/3)[argsh(t^(3/2) )]_1 ^x =[(2/3)ln∣(√t^3 )+(√(1+t^3 ))∣]_1 ^x =(2/3)ln((√x^3 )+(√(1+x^3 )))−(2/3)ln(1+(√2))](https://www.tinkutara.com/question/Q191656.png)
$$\int_{\mathrm{1}} ^{{x}} \sqrt{\frac{{t}}{\mathrm{1}+{t}^{\mathrm{3}} }}{dt} \\ $$$$=\int_{\mathrm{1}} ^{{x}} \frac{{t}^{\frac{\mathrm{1}}{\mathrm{2}}} }{\:\sqrt{\mathrm{1}+\left({t}^{\frac{\mathrm{3}}{\mathrm{2}}} \right)^{\mathrm{2}} }}{dt}=\frac{\mathrm{2}}{\mathrm{3}}\int_{\mathrm{1}} ^{{x}} \frac{{d}\left({t}^{\frac{\mathrm{3}}{\mathrm{2}}} \right)}{\:\sqrt{\mathrm{1}+\left({t}^{\frac{\mathrm{3}}{\mathrm{2}}} \right)^{\mathrm{2}} }} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\left[\mathrm{argsh}\left({t}^{\frac{\mathrm{3}}{\mathrm{2}}} \right)\right]_{\mathrm{1}} ^{{x}} =\left[\frac{\mathrm{2}}{\mathrm{3}}\mathrm{ln}\mid\sqrt{{t}^{\mathrm{3}} }+\sqrt{\mathrm{1}+{t}^{\mathrm{3}} }\mid\right]_{\mathrm{1}} ^{{x}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\mathrm{ln}\left(\sqrt{{x}^{\mathrm{3}} }+\sqrt{\mathrm{1}+{x}^{\mathrm{3}} }\right)−\frac{\mathrm{2}}{\mathrm{3}}\mathrm{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$