How do you integrate #int x/sqrt(x+1)# using substitution?

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Answer 1 · 2016-10-24T12:38:53

#y=(2(x+1)^(3/2))/3-2sqrt(x+1)+c#

so let us take,

#u=x+1#

so we differentiate this to find that

#(du)/(dx)=1#

and #x=u-1#

so let's rewrite our original integral

#int(x/sqrt(x+1))dx#

#=int((u-1)/sqrt(u))(1)dx#

#=int(u/sqrt(u)-1/sqrt(u))(du)/(dx)dx#

#=int(sqrt(u)-1/sqrt(u))(du)#

#=(2u^(3/2))/3-2sqrt(u)+c#

we can now sub in the #x+1 " for the u's"# if we need,

#=(2(x+1)^(3/2))/3-2sqrt(x+1)+c#

Answer 2 · 2016-10-24T12:50:46

The integral is #=2/3sqrt(x+1)(x-2)+C#

Let #u=x+1# and #x=u-1#

the #du=dx#
So#int(xdx)/sqrt(x+1)=int((u-1)du)/u^(1/2)#

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

#=u^(3/2)/(3/2)-u^(1/2)/(1/2)#

#=2/3u*sqrtu-2sqrtu#
#=2*sqrtu(1/3u-1)#
#=2/3sqrtu(u-3)#

Replacing u by (x+1)

#int(xdx)/sqrt(x+1)=2/3sqrt(x+1)(x+1-3)+C#

#=2/3sqrt(x+1)(x-2)+C#