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 , $u=sqrt(2x-1)$

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

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