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How do you integrate #int xsqrt(2x+1)# using substitution?

1 Answer

Jul 26, 2016

#= 1/10(2x+1)^(5/2) - 1/6(2x+1)^(3/2) + C#

Explanation:

Let #u = 2x+1 implies du = 2dx#

#implies x = (u-1)/2 and dx = (du)/2#

#int xsqrt(2x+1)dx = 1/4 int (u-1)u^(1/2)du#

#= 1/4int u^(3/2) - u^(1/2)du#

#=1/4[2/5u^(5/2) - 2/3u^(3/2)] + C#

#= 1/10(2x+1)^(5/2) - 1/6(2x+1)^(3/2) + C#

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