How do you integrate ((x^2-1)/sqrt(2x-1) )dx(x212x1)dx?

1 Answer
May 28, 2018

I=1/60[3(sqrt(2x-1))^5+10(sqrt(2x-1))^3-45sqrt(2x-1)]+cI=160[3(2x1)5+10(2x1)3452x1]+c

Explanation:

Here,

I=int((x^2-1)/sqrt(2x-1))dxI=(x212x1)dx

Subst, sqrt(2x-1)=u=>2x-1=u^2=>2x=u^2+12x1=u2x1=u22x=u2+1

=>x=1/2(u^2+1)=>dx=1/2(2u)du=udux=12(u2+1)dx=12(2u)du=udu

So,

I=int(1/4(u^2+1)^2-1)/uxxuduI=14(u2+1)21u×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+14]du

=1/4int[u^4+2u^2-3]du=14[u4+2u23]du

=1/4[u^5/5+2u^3/3-3u]+c=14[u55+2u333u]+c

=1/4xx1/15[3u^5+10u^3-45u]+c=14×115[3u5+10u345u]+c

=1/60[3(u)^5+10(u)^3-45u]+c=160[3(u)5+10(u)345u]+c

Subst, back , u=sqrt(2x-1)u=2x1

I=1/60[3(sqrt(2x-1))^5+10(sqrt(2x-1))^3-45sqrt(2x-1)]+cI=160[3(2x1)5+10(2x1)3452x1]+c