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 108663 by bobhans last updated on 18/Aug/20

Answered by john santu last updated on 18/Aug/20

     (((⊸J)/(S⇋))/)  let (√(1+x^3 )) = m ⇒x^3  = m^2 −1  ⇒3x^2  dx = 2m dm  I= ∫ ((x^3 .x^2 )/( (√(1+x^3 )))) dx = (2/3)∫ (((m^2 −1)m dm)/m)  = (2/3)∫ (m^2 −1) dm   = (2/9)m^3 −(2/3)m + C  = (2/9)(1+x^3 )−(2/3)(√(1+x^3 )) + C

$$\:\:\:\:\:\frac{\frac{\multimap{J}}{{S}\leftrightharpoons}}{} \\ $$$${let}\:\sqrt{\mathrm{1}+{x}^{\mathrm{3}} }\:=\:{m}\:\Rightarrow{x}^{\mathrm{3}} \:=\:{m}^{\mathrm{2}} −\mathrm{1} \\ $$$$\Rightarrow\mathrm{3}{x}^{\mathrm{2}} \:{dx}\:=\:\mathrm{2}{m}\:{dm} \\ $$$${I}=\:\int\:\frac{{x}^{\mathrm{3}} .{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}+{x}^{\mathrm{3}} }}\:{dx}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\int\:\frac{\left({m}^{\mathrm{2}} −\mathrm{1}\right){m}\:{dm}}{{m}} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{3}}\int\:\left({m}^{\mathrm{2}} −\mathrm{1}\right)\:{dm}\: \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{9}}{m}^{\mathrm{3}} −\frac{\mathrm{2}}{\mathrm{3}}{m}\:+\:{C} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{9}}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)−\frac{\mathrm{2}}{\mathrm{3}}\sqrt{\mathrm{1}+{x}^{\mathrm{3}} }\:+\:{C} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com