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## Homework Statement

[tex]\int \frac{1}{1+\sqrt{2x}}dx[/tex]

## Homework Equations

[tex]u=1+\sqrt{2x}[/tex]

[tex]\sqrt{2x}=u-1[/tex]

[tex]dx=(u-1)du[/tex]

## The Attempt at a Solution

I was able to get it down to:

[tex]\int (1-\frac{1}{u})du[/tex]

[tex]= u-\ln{lul}}+C[/tex]

[tex]= 1+\sqrt{2x}-\ln{l1+\sqrt{2x}l}+C[/tex]

However, my book says that the solution to the integral is:

[tex] \sqrt{2x}-\ln{l{1+\sqrt{2x}l}+C[/tex] (Without the 1 in front)

Why is this? Thanks in advance for your help!

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