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 11149 by suci last updated on 14/Mar/17

∫x^5 ((x^3 +1))^(1/3) dx=...???

$$\int{x}^{\mathrm{5}} \:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} +\mathrm{1}}\:{dx}=...??? \\ $$

Answered by ajfour last updated on 14/Mar/17

(((x^3 +1)^(7/3) )/7)−(((x^3 +1)^(4/3) )/4) +C

$$\:\frac{\left({x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{7}/\mathrm{3}} }{\mathrm{7}}−\frac{\left({x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{4}/\mathrm{3}} }{\mathrm{4}}\:+{C}\: \\ $$

Commented by ajfour last updated on 14/Mar/17

let x^3 +1=t 3x^2 dx=dt so, ∫x^5 (x^3 +1)^(1/3) dx = (1/3)∫x^3 (x^3 +1)^(1/3) 3x^2 dx =(1/3)∫(t−1)t^(1/3) dt=(1/3)∫(t^(4/3) −t^(1/3) )dt = (1/3)((t^(7/3) /(7/3)))−(1/3)((t^(4/3) /(4/3))) +C hence the answer.

$${let}\:\:\:{x}^{\mathrm{3}} +\mathrm{1}={t} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} {dx}={dt} \\ $$$${so},\:\int{x}^{\mathrm{5}} \left({x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{1}/\mathrm{3}} {dx}\: \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{3}}\int{x}^{\mathrm{3}} \left({x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{1}/\mathrm{3}} \mathrm{3}{x}^{\mathrm{2}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int\left({t}−\mathrm{1}\right){t}^{\mathrm{1}/\mathrm{3}} {dt}=\frac{\mathrm{1}}{\mathrm{3}}\int\left({t}^{\mathrm{4}/\mathrm{3}} −{t}^{\mathrm{1}/\mathrm{3}} \right){dt} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{{t}^{\mathrm{7}/\mathrm{3}} }{\mathrm{7}/\mathrm{3}}\right)−\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{{t}^{\mathrm{4}/\mathrm{3}} }{\mathrm{4}/\mathrm{3}}\right)\:+{C} \\ $$$${hence}\:{the}\:{answer}. \\ $$

Answered by Mechas88 last updated on 17/Mar/17

u=(x^3 +1)^(1/3) u^3 =x^3 +1 x^3 =u^3 −1 du=(1/3)(x^3 +1)^(−2/3) (3x^2 )dx=x^2 (x^3 +1)^(−2/3) dx dx=x^(−2) (x^3 +1)^(2/3) du ∫x^5 x^(−2) udu=∫(u^3 −1)udu= ∫u^4 du−∫udu= (u^5 /5) − (u^2 /2) + C= (((((x^3 +1)^5 ))^(1/3) )/5) −((((x^3 +1)^2 ))^(1/3) /2) + C ∗∗∗Rta

$$ \\ $$$${u}=\left({x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$${u}^{\mathrm{3}} ={x}^{\mathrm{3}} +\mathrm{1} \\ $$$${x}^{\mathrm{3}} ={u}^{\mathrm{3}} −\mathrm{1} \\ $$$${du}=\frac{\mathrm{1}}{\mathrm{3}}\left({x}^{\mathrm{3}} +\mathrm{1}\right)^{−\mathrm{2}/\mathrm{3}} \left(\mathrm{3}{x}^{\mathrm{2}} \right){dx}={x}^{\mathrm{2}} \left({x}^{\mathrm{3}} +\mathrm{1}\right)^{−\mathrm{2}/\mathrm{3}} {dx} \\ $$$${dx}={x}^{−\mathrm{2}} \left({x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{2}/\mathrm{3}} {du} \\ $$$$ \\ $$$$\int{x}^{\mathrm{5}} {x}^{−\mathrm{2}} {udu}=\int\left({u}^{\mathrm{3}} −\mathrm{1}\right){udu}= \\ $$$$\int{u}^{\mathrm{4}} {du}−\int{udu}= \\ $$$$\frac{{u}^{\mathrm{5}} }{\mathrm{5}}\:−\:\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\:+\:{C}= \\ $$$$\frac{\sqrt[{\mathrm{3}}]{\left({x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{5}} }\:}{\mathrm{5}}\:−\frac{\sqrt[{\mathrm{3}}]{\left({x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{2}} \:\:\:\:}}{\mathrm{2}}\:\:+\:\:{C}\:\:\ast\ast\ast{Rta} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com