I dont understand why my integration goes wrong. Need help

Question

I have the following that i need to integrate: $$ \int \ln(1 + \sqrt{x}) \, \mathrm{d}x $$

1. I use substitution first: $u = 1 + \sqrt{x}$, which gives: $2\sqrt{x}\,\mathrm{d}u = \mathrm{d}x$

2. Thereafter, I do integration by parts on $\ln x$, which gives: $x\cdot\ln x - x$

3. Then I go back to substitution again, and insert the answer from 1: $2\sqrt{x}\,\mathrm{d}u = \mathrm{d}x$: $$ \int \ln(1 + \sqrt{x})\,\mathrm{d}x = \ln(u) \cdot 2\sqrt{x}\,\mathrm{d}u = x\cdot\ln u - x\cdot 2\sqrt{x} $$

4. I insert back $u$ into the result in 3: $u = 1 + \sqrt{x}$

5. Result: $x\cdot\ln(1 + \sqrt{x}) - x\cdot2\sqrt{x}$

But this is wrong, why?

Answer 1

$u$ and $x$ are not independent variables, so this step is wrong:

$$\int \ln(u) \cdot 2\sqrt{x} \mathrm{d}u = x\cdot\ln u - x\cdot 2\sqrt{x}$$

It is in general advisable to replace every occurrence of $x$ with the new variable (in this case $u$) when integration by substitution is used.

Answer 2

Using $u=1+\sqrt{x}$ you get

\begin{eqnarray} \int \ln(1+\sqrt{x})\,dx&=&\int 2(u-1)\ln u\,du\\ &=&\int \ln u\,d(u-1)^2\\ &=& (u-1)^2\ln u-\int(u-1)^2\,d\ln u\\ &=&x\ln(1+\sqrt{x})-\int\frac{u^2-2u+1}{u}\,du\\ &=&x\ln(1+\sqrt{x})-\int u-2+\frac{1}{u}\,du\\ &=&x\ln(1+\sqrt{x})-\frac{u^2}{2}-2u+\ln u +c\\ &=&x\ln(1+\sqrt{x})-\frac{x}{2}-2\sqrt{x}+\ln(1+\sqrt{x}) +c \end{eqnarray}

Answer 3

Thanks everyone! So i finally got the right answer: 1/2u^2ln(u)-uln(u)-1/4u^2+u.

And when inserting u=1+√x: 1/2(1+√x)^2*ln(1+√x)-(1+√x)*ln(1+√x)-1/4(1+√x)^2+(1+√x).

Gives also the right result of 1/2=0.5, when inserting the limits of limits 0 to 1.