Question #6f533

1 Answer
sente
Dec 15, 2016

#intxsqrt(2x+1)dx = (2x+1)^(3/2)(1/5x-1/15)+C#

Explanation:

Using the integration by parts formula

#intudv = uv-intvdu#

Let
#u = x => du = dx#
#dv = (2x+1)^(1/2)dx => v = 1/3(2x+1)^(3/2)#

#intxsqrt(2x+1)dx = intudv#

#=uv - intvdu#

#=1/3x(2x+1)^(3/2)-1/3int(2x+1)^(3/2)dx#

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

#=1/3x(2x+1)^(3/2)-1/15(2x+1)^(5/2)+C#

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

#=(2x+1)^(3/2)(1/5x-1/15)+C#

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