How do you integrate #int sqrt(4x-5)dx# using substitution?

Question

5808 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-18T15:42:35

Let #u = 4x - 5#, then #(du)/dx = 4 ->dx = (du)/4#

#=1/4int(sqrt(u))#

#=1/4(u^(1/2 + 1)/(1/2 + 1)) + C#

#=1/4(2/3u^(3/2)) + C#

#=1/4(2/3(4x -5)^(3/2)) + C#

#=1/6(4x - 5)^(3/2) + C#

Hopefully this helps!