How do you integrate #((x^2-1)/sqrt(2x-1) )dx#?

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Answer 1 · 2018-05-28T11:11:53

#I=1/60[3(sqrt(2x-1))^5+10(sqrt(2x-1))^3-45sqrt(2x-1)]+c#

Here,

#I=int((x^2-1)/sqrt(2x-1))dx#

Subst, #sqrt(2x-1)=u=>2x-1=u^2=>2x=u^2+1#

#=>x=1/2(u^2+1)=>dx=1/2(2u)du=udu#

So,

#I=int(1/4(u^2+1)^2-1)/uxxudu#

#I=int[1/4(u^4+2u^2+1)-1]du#

#=1/4int[u^4+2u^2+1-4]du#

#=1/4int[u^4+2u^2-3]du#

#=1/4[u^5/5+2u^3/3-3u]+c#

#=1/4xx1/15[3u^5+10u^3-45u]+c#

#=1/60[3(u)^5+10(u)^3-45u]+c#

Subst, back , #u=sqrt(2x-1)#

#I=1/60[3(sqrt(2x-1))^5+10(sqrt(2x-1))^3-45sqrt(2x-1)]+c#