Integration of a natural logarithmic function using substitution

Question

The function is $$\int \frac{1}{1+ \sqrt{2x}} dx$$

Using $u=1+\sqrt{2x}$, $du=\frac{1}{\sqrt{2x}}dx$. Which gives us $$\sqrt{2x}\int\frac{1}{u}du $$ The final answer of this would give $\sqrt{2x}\ln|1+\sqrt{2x}|$, but when I checked my answer the correct answer is $\sqrt{2x}-\ln|1+\sqrt{2x}|$.

Answer 1

The way you have solved the problem is wrong. When you make the substitution

$$u = 1+\sqrt{2x}$$

$$\implies x = \frac{(u-1)^2}{2}$$ $$\implies dx = (u-1)du$$

Therefore the integral then is:

$$\int \frac{u-1}{u} du$$ $$= u - \log(|u|) + C$$ $$= 1+\sqrt{2x} - \log(|1+\sqrt{2x}|) +C$$ $$=\sqrt{2x} - \log(|1+\sqrt{2x}|) +C'$$

Answer 2

Let $u=1+\sqrt{2x}$, then $$x=\frac{1}{2}(u-1)^2$$ and $$dx=(u-1)du$$ Therefore $$\begin{align}\int \frac{1}{1+\sqrt{2x}}dx&=\int \frac{1}{u}(u-1)du \\ &=\int \left( 1-\frac{1}{u}\right)du \\ &=u-\log |u|+c \\ &=(1+\sqrt{2x})-\log(1+\sqrt{2x})+c\end{align}$$