integral with several answers Could you help me please with a question about integrals? Can an integral have more than one answer? For example with this integral: 
$$\int\sqrt{1+\sqrt{1-x^2}}dx$$
Doing by replacing u=$\sqrt{1-x^2}$, I have this solution: 
$$2\sqrt{1-\sqrt{1-x^2}}-\frac23 \left(1-\sqrt{1-x^2} \right)^{3/2},$$
and another solution doing by replacing $\sin \theta =x/1$, I get:
$$2\sqrt{2} \sin(\arcsin(x/2))-\frac43 \sqrt{2} \sin^3(\arcsin(x/2))).$$
The graph of each one is in this picture:

Thanks in advance.
The error in one of your solutions lies in the fact that you have to consider the sign of the $x$ values used when performing the integration. In particular, using $u=\sqrt{1-x^2}$, we cannot say that $x=\sqrt{1-u^2}$ because $x$ may be negative whereas the square root term cannot. Also I believe you forgot to use the fact that $\sqrt{x^2}=|x|$ in your working. The trigonometric substitution has given you the correct answer for the integral in the interval $x\in[-1,1]$. We can get the correct algebraic solution by using the identity
$$\sin{\left(\frac{\arcsin{(x)}}2\right)}=\text{sgn}(x)\sqrt{\frac{1-\sqrt{1-x^2}}2}$$
for $x\in[-1,1]$. Here $\text{sgn}(x)$ denotes the Sign function. This gives the value of the integral as
$$2\text{sgn}(x)\sqrt{1-\sqrt{1-x^2}}-\frac23\text{sgn}(x) (1-\sqrt{1-x^2})^{3/2}+C$$
instead of your solution. The two answers are then the exact same function.
Yes, an integral can be have more than one "answer", hence the importance of the constant that is added when finished integrating. More correctly, if two functions $f$ and $g$ they have the same derivative, so we can say that these differ by a constant :
$$f'=g' \, \Rightarrow \, f=g+c \, \textrm{ for some } c$$
Think of a simpler example, let
$$\int 2 \sin x \cos x \ dx$$
then, if we take $u=\sin x$, this integral is equal to $\sin^2 x$ and if we take $u=\cos x$ the integral is $-\cos^2 x$.

As you can see, the integral has two (or more) answers, and that clearly $\sin^2 x \neq -\cos^2 x$.

In fact, this two functions differ by a constant, by $1$ :
$$\sin^2 x = -\cos^2 x \color{red}{+1}$$