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 40625 by Raj Singh last updated on 25/Jul/18

Answered by MJS last updated on 25/Jul/18

∫(dx/((√x)+(x)^(1/3) ))= [t=(x)^(1/6) → dx=6(x^5 )^(1/6) dt] =6∫(t^3 /(t+1))dt=6∫(t^2 −t+1−(1/(t+1)))dt= =6((1/3)t^3 −(1/2)t^2 +t−ln (t+1))= =2t^3 −3t^2 +6t−6ln (t+1)= =2(√x)−3(x)^(1/3) +6(x)^(1/6) −6ln (1+(x)^(1/6) ) +C

$$\int\frac{{dx}}{\sqrt{{x}}+\sqrt[{\mathrm{3}}]{{x}}}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt[{\mathrm{6}}]{{x}}\:\rightarrow\:{dx}=\mathrm{6}\sqrt[{\mathrm{6}}]{{x}^{\mathrm{5}} }{dt}\right] \\ $$$$=\mathrm{6}\int\frac{{t}^{\mathrm{3}} }{{t}+\mathrm{1}}{dt}=\mathrm{6}\int\left({t}^{\mathrm{2}} −{t}+\mathrm{1}−\frac{\mathrm{1}}{{t}+\mathrm{1}}\right){dt}= \\ $$$$=\mathrm{6}\left(\frac{\mathrm{1}}{\mathrm{3}}{t}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} +{t}−\mathrm{ln}\:\left({t}+\mathrm{1}\right)\right)= \\ $$$$=\mathrm{2}{t}^{\mathrm{3}} −\mathrm{3}{t}^{\mathrm{2}} +\mathrm{6}{t}−\mathrm{6ln}\:\left({t}+\mathrm{1}\right)= \\ $$$$=\mathrm{2}\sqrt{{x}}−\mathrm{3}\sqrt[{\mathrm{3}}]{{x}}+\mathrm{6}\sqrt[{\mathrm{6}}]{{x}}−\mathrm{6ln}\:\left(\mathrm{1}+\sqrt[{\mathrm{6}}]{{x}}\right)\:+{C} \\ $$

Commented by MJS last updated on 25/Jul/18

...now try ∫(dx/((√x)+(x)^(1/5) )) and ∫(dx/((x)^(1/3) +(x)^(1/5) ))

$$...\mathrm{now}\:\mathrm{try} \\ $$$$\int\frac{{dx}}{\sqrt{{x}}+\sqrt[{\mathrm{5}}]{{x}}}\:\mathrm{and}\:\int\frac{{dx}}{\sqrt[{\mathrm{3}}]{{x}}+\sqrt[{\mathrm{5}}]{{x}}} \\ $$

Answered by Joel578 last updated on 25/Jul/18

x = t^6 → dx = 6t^5 dt I = ∫ ((6t^5 )/(t^3 + t^2 )) dt = ∫ ((6t^3 )/(t + 1)) dt = ∫ (6t^2 − 6t − (6/(t + 1)) + 6) dt = 2t^3 − 3t^2 − 6 ln ∣t + 1∣ + 6t + C = 2x^(1/2) − 3x^(1/3) − 6 ln ∣x^(1/6) + 1∣ + 6x^(1/6) + C

$${x}\:=\:{t}^{\mathrm{6}} \:\:\rightarrow\:\:{dx}\:=\:\mathrm{6}{t}^{\mathrm{5}} \:{dt} \\ $$$${I}\:=\:\int\:\frac{\mathrm{6}{t}^{\mathrm{5}} }{{t}^{\mathrm{3}} \:+\:{t}^{\mathrm{2}} }\:{dt}\:=\:\int\:\frac{\mathrm{6}{t}^{\mathrm{3}} }{{t}\:+\:\mathrm{1}}\:{dt} \\ $$$$\:\:\:\:=\:\int\:\left(\mathrm{6}{t}^{\mathrm{2}} \:−\:\mathrm{6}{t}\:−\:\frac{\mathrm{6}}{{t}\:+\:\mathrm{1}}\:+\:\mathrm{6}\right)\:{dt} \\ $$$$\:\:\:\:=\:\mathrm{2}{t}^{\mathrm{3}} \:−\:\mathrm{3}{t}^{\mathrm{2}} \:−\:\mathrm{6}\:\mathrm{ln}\:\mid{t}\:+\:\mathrm{1}\mid\:+\:\mathrm{6}{t}\:+\:{C} \\ $$$$\:\:\:\:=\:\mathrm{2}{x}^{\frac{\mathrm{1}}{\mathrm{2}}} \:−\:\mathrm{3}{x}^{\frac{\mathrm{1}}{\mathrm{3}}} \:−\:\mathrm{6}\:\mathrm{ln}\:\mid{x}^{\frac{\mathrm{1}}{\mathrm{6}}} \:+\:\mathrm{1}\mid\:+\:\mathrm{6}{x}^{\frac{\mathrm{1}}{\mathrm{6}}} \:+\:{C} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com