Here,
I=int((x^2-1)/sqrt(2x-1))dxI=∫(x2−1√2x−1)dx
Subst, sqrt(2x-1)=u=>2x-1=u^2=>2x=u^2+1√2x−1=u⇒2x−1=u2⇒2x=u2+1
=>x=1/2(u^2+1)=>dx=1/2(2u)du=udu⇒x=12(u2+1)⇒dx=12(2u)du=udu
So,
I=int(1/4(u^2+1)^2-1)/uxxuduI=∫14(u2+1)2−1u×udu
I=int[1/4(u^4+2u^2+1)-1]duI=∫[14(u4+2u2+1)−1]du
=1/4int[u^4+2u^2+1-4]du=14∫[u4+2u2+1−4]du
=1/4int[u^4+2u^2-3]du=14∫[u4+2u2−3]du
=1/4[u^5/5+2u^3/3-3u]+c=14[u55+2u33−3u]+c
=1/4xx1/15[3u^5+10u^3-45u]+c=14×115[3u5+10u3−45u]+c
=1/60[3(u)^5+10(u)^3-45u]+c=160[3(u)5+10(u)3−45u]+c
Subst, back , u=sqrt(2x-1)u=√2x−1
I=1/60[3(sqrt(2x-1))^5+10(sqrt(2x-1))^3-45sqrt(2x-1)]+cI=160[3(√2x−1)5+10(√2x−1)3−45√2x−1]+c