Evaluation of $\int_{0}^{a} \frac{\sqrt{a+x}}{\sqrt{a-x}} dx$ Evaluate : 
$\int_{0}^{a} \frac{\sqrt{a+x}}{\sqrt{a-x}} dx$ 
My approach : I multiplied both sides by $\sqrt{a+x}$ and after simplification it comes down to : 
$\int_{0}^{a} \frac{a}{\sqrt{a^{2}-x^{2}}} dx + \int_{0}^{a} \frac{x}{\sqrt{a^{2}-x^{2}}} dx$, let's denote them by $I_1$ and $I_2$ respectively.
$I_1$ can be easily solved to $a \sin ^{-1} \frac{x}{a}$. 
If I write $I_2$ as $ \frac{-1}{2}\int_{0}^{a} \frac{-2x}{\sqrt{a^{2}-x^{2}}} dx$, I get the solution as $\frac{a}{2}(\pi+1)$ but the answer given is $\frac{a}{2}(\pi+2)$. Where did I go wrong?

So for your solution: 
$$I_2 = \int_{0}^{a} \frac{x}{\sqrt{a^2 - x^2}} dx$$
Let $u = a^2 - x^2, du = -2x dx, dx = -\frac{1}{2x} du$ then it is clear that 
$$ I_2 = - \frac{1}{2} \int \frac{1}{\sqrt{u}} du = -  \sqrt{u} $$
Now evaluating back at $x$, 
$$ = - \sqrt{a^2 - x^2} | _0^a = - \sqrt{0} + \sqrt{a^2} = a$$
I would recommend against, the expression
$$ - \frac{1}{2} \int_{0}^{a} \frac{-2x}{\sqrt{a^2 - x^2 }} dx $$ 
Since it doesn't really make it any easier to integrate.But if we insist, again making the substitution $u = a^2 - x^2, dx = -\frac{1}{2x} du$ we get:
$$ - \frac{1}{2} \int \frac{1}{\sqrt{u}} dx $$ 
Which is evaluated the same way.
