Answer to Question #50467 in Integral Calculus for TINU

Question #50467

what is the integration of (square root of x)/(1+(cube root of x))

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20.01.15, 17:06

Dear tinu. To get the third line, it is necessary to perform several operations over the integrand, namely, u[sup]8[/sup]/(1+u[sup]2[/sup])=(u[sup]8[/sup]+u[sup]6[/sup]-u[sup]6[/sup]-u[sup]4[/sup]+u[sup]4[/sup]+u[sup]2[/sup]-u[sup]2[/sup]-1+1)/(1+u[sup]2[/sup])=|additive property|=(u[sup]8[/sup]+u[sup]6[/sup])/(1+u[sup]2[/sup])-(u[sup]6[/sup]+u[sup]4[/sup])/(1+u[sup]2[/sup])+(u[sup]4[/sup]+u[sup]2[/sup])/(1+u[sup]2[/sup])-(u[sup]2[/sup]+1)/(1+u[sup]2[/sup])+1/(1+u[sup]2[/sup])=u[sup]6[/sup](u[sup]2[/sup]+1)/(1+u[sup]2[/sup])-u[sup]4[/sup](u[sup]2[/sup]+1)/(1+u[sup]2[/sup])+ u[sup]2[/sup](u[sup]2[/sup]+1)/(1+u[sup]2[/sup])-1+1/(1+u[sup]2[/sup])= u[sup]6[/sup]-u[sup]4[/sup]+ u[sup]2[/sup]-1+1/(1+u[sup]2[/sup]). Thank you for using our service.

tinu

20.01.15, 13:07

i am unable to get the 3rd line ? can you explain how did you break u^8??

Answer to Question #50467 in Integral Calculus for TINU

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