How do you find the integral of #dx/((x^(5/2) * (1+ x^2)^(1/4))#?

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Answer 1 · 2018-07-21T17:35:42

#int1/(x^(5/2)(1+x^2)^(1/4))dx=-2/3(1/x^2+1)^(3/4)+c#

Here ,

#I=int1/(x^(5/2)(1+x^2)^(1/4))dx#

Substitute #color(red)(x=1/t=>dx=-1/t^2dt#

So,

#I=int1/((1/t)^(5/2)(1+1/t^2)^(1/4))(-1/t^2)dt#

#=-intt^(5/2)/((t^2+1)^(1/4)/t^(1/2))*1/t^2dt#

#=-intt^(5/2+1/2-2)/(t^2+1)^(1/4)dt#

#:.I=-intt/(t^2+1)^(1/4)dt#

Substitute #color(blue)((t^2+1)^(1/4)=u=>t^2+1=u^4#

#=>2tdt=4u^3du=>tdt=2u^3du #

#I=-int(2u^3)/udu#

#=>I=-2intu^2du#

#=>I=-2[u^3/3]+c#

#=-2/3(u)^3+c#

Subst. back , #color(blue)(u=(t^2+1)^(1/4)#

#I=-2/3(t^2+1)^(3/4)+c#

Again subst. #color(red)(t=1/x#

#:.I=-2/3(1/x^2+1)^(3/4)+c#

#=>I=-2/3(1+x^2)^(3/4)/(x^2)^(3/4)+c#

#:.I=-(2(1+x^2)^(3/4))/(3x^(3/2))+c#