Error in solving $\int \sqrt{1 + e^x} dx$ . I want to solve this integral for $1 + e^x \ge 0$
$$\int \sqrt{1 + e^x} dx$$ 
I start by parts $$\int \sqrt{1 + e^x} dx = x\sqrt{1 + e^x} - \int x \frac{e^x}{2\sqrt{1 + e^x}} dx $$
Substitute $\sqrt{1 + e^x} = t \implies dt = \frac{e^x}{2\sqrt{1 + e^x}} dx$
So I remain with $$x\sqrt{1 + e^x} - \int \ln((t-1)(t+1)) dt =  x\sqrt{1 + e^x}  + 2t +(-t+1)\ln(t-1)+ (-t+1)\ln(t+1) $$
that is
$$(x+2)\sqrt{1 + e^x} + (-\sqrt{1 + e^x} + 1) \ln(+\sqrt{1 + e^x} - 1) + (-\sqrt{1 + e^x} + 1) \ln(+\sqrt{1 + e^x} + 1) + C  $$
But the solution should be $$2\sqrt{1 + e^x} + \ln(+\sqrt{1 + e^x} - 1) - \ln(+\sqrt{1 + e^x} + 1) $$
Where is the mistake hidden?

HINT:
Let $\sqrt{1+e^x}=y\implies e^x=y^2-1\implies e^x\ dx=2y\ dy$
$$\int\sqrt{1+e^x}\ dx=\int\dfrac{2y^2}{y^2-1}dy$$
Now $y^2=y^2-1+1$
$$\dfrac{2y^2}{y^2-1}=2+\dfrac{y+1-(y-1)}{y^2-1}=2+\dfrac1{y-1}-\dfrac1{y+1}$$
