Find the indefinite integral by u-sub

[InaudibleTree]

Homework Statement

\int1/(1+\sqrt{2x})\,dx

Homework Equations

u=1+\sqrt{2x} \Rightarrow \sqrt{2x}=u-1
du=1/\sqrt{2x}dx \Rightarrow \sqrt{2x}du=dx

The Attempt at a Solution

\int1/(1+\sqrt{2x})\,dx = \int\sqrt{2x}/(1+\sqrt{2x})\,du = \int(u-1)/u\,du = \int\,du-\int1/u\,du = u-ln|u|+C = 1+\sqrt{2x}-ln|1+\sqrt{2x}|+C

The book I am using has the answer as:
\sqrt{2x}-ln|1+\sqrt{2x}|+C

Where am i going wrong?

 

[Curious3141]

Homework Statement

\int1/(1+\sqrt{2x})\,dx

Homework Equations

u=1+\sqrt{2x} \Rightarrow \sqrt{2x}=u-1
du=1/\sqrt{2x}dx \Rightarrow \sqrt{2x}du=dx

The Attempt at a Solution

\int1/(1+\sqrt{2x})\,dx = \int\sqrt{2x}/(1+\sqrt{2x})\,du = \int(u-1)/u\,du = \int\,du-\int1/u\,du = u-ln|u|+C = 1+\sqrt{2x}-ln|1+\sqrt{2x}|+C

The book I am using has the answer as:
\sqrt{2x}-ln|1+\sqrt{2x}|+C

Where am i going wrong?

That "extra" 1 in your answer can be "absorbed" into the arbitrary constant C (so you can drop the 1). The answers are equivalent.

 

[InaudibleTree]

That "extra" 1 in your answer can be "absorbed" into the arbitrary constant C (so you can drop the 1). The answers are equivalent.

Oh ok. Thank you curious.