How do you integrate #((x^2-1)/sqrt(2x-1) )dx#? Calculus Techniques of Integration Integration by Substitution 1 Answer maganbhai P. May 28, 2018 #I=1/60[3(sqrt(2x-1))^5+10(sqrt(2x-1))^3-45sqrt(2x-1)]+c# Explanation: 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# Answer link Related questions What is Integration by Substitution? How is integration by substitution related to the chain rule? How do you know When to use integration by substitution? How do you use Integration by Substitution to find #intx^2*sqrt(x^3+1)dx#? How do you use Integration by Substitution to find #intdx/(1-6x)^4dx#? How do you use Integration by Substitution to find #intcos^3(x)*sin(x)dx#? How do you use Integration by Substitution to find #intx*sin(x^2)dx#? How do you use Integration by Substitution to find #intdx/(5-3x)#? How do you use Integration by Substitution to find #intx/(x^2+1)dx#? How do you use Integration by Substitution to find #inte^x*cos(e^x)dx#? See all questions in Integration by Substitution Impact of this question 8092 views around the world You can reuse this answer Creative Commons License