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.

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

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

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