How to integrate #int12x[sqrt(x+1)]dx# using integration by parts ?

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Answer 1 · 2018-04-30T14:46:16

The answer is #=8x(1+x)^(3/2)-16/5(1+x)^(5/2)+C#

We need

We solve this integral by integration by parts

#intuv'dx=uv-intu'v#

Here,

#u=(12x)#, #=>#, #u'=12#

#v'=sqrt(x+1)#, #=>#, #v=2/3(x+1)^(3/2)#

Therefore,

#int12xsqrt(x+1)dx=12x*2/3(1+x)^(3/2)-int12*2/3(1+x)^(3/2)dx#

#=8x(1+x)^(3/2)-8*2/5*(1+x)^(5/2)+C#

#=8x(1+x)^(3/2)-16/5(1+x)^(5/2)+C#