How do you evaluate the integral #int sqrt(2x+3)dx#?

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Answer 1 · 2017-01-09T03:46:21

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

Let #u = 2x + 3#. Then #du = 2dx# and #dx= 1/2du#

#=1/2intsqrt(u)du#

#=1/2int u^(1/2)du#

You can integrate this as #int(x^n)dx = x^(n + 1)/(n + 1) + C#:

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

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

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

Hopefully this helps!