How do you find the integral for #(x)/(sqrt(1+2x))dx#?

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Answer 1 · 2015-04-01T14:50:13

I would probably use parts, but since this is asked in Integration by substitution, I'll do it that way.

#int (x)/(sqrt(1+2x))dx#

Let #u = 1+2x#. This makes #du = 2 dx# and #dx= 1/2 du#.

The substitution is not complete until we also note that
#u = 1+2x# makes #x= (u-1)/2#

#int (x)/(sqrt(1+2x))dx = int x 1/(sqrt(1+2x)) *dx #

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

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

You can probably finish this yourself.

# = 1/4[2/3 u^(3/2) - 2/1 u^(1/2)]+C = 1/6[u^(3/2)-3u^(1/2)]+C#

#int (x)/(sqrt(1+2x))dx =1/6[sqrt(1+2x)^3-3sqrt(1+2x)]+C#