So cute and yummy!

Note first that this is an improper integral (at the upper limit), so for now we will just deal with the indefinite integral and handle the limits at the end.

Multiply both the numerator and denominator of the integrand by $\sqrt{1 + x}.$ The (indefinite) integral then becomes

$\displaystyle\int \dfrac{1 + x}{\sqrt{1 - x^{2}}} dx = \int \dfrac{1}{\sqrt{1 - x^{2}}} dx + \int \dfrac{x}{\sqrt{1 - x^{2}}} dx.$

For the first of these integrals let $x = \sin(\theta).$ Then $dx = \cos(\theta) d\theta$ and we just end up with

$\displaystyle \int \dfrac{\cos(\theta)}{\sqrt{1 - \sin^{2}(\theta)}} d\theta = \theta = \sin^{-1}(x),$

(where I've ignored the constant of integration as we will make this integral definite in the last step).

For the second integral we use the substitution $u = 1 - x^{2}.$ Then $du = -2x dx,$ and the integral then becomes

$\displaystyle -\dfrac{1}{2}\int \dfrac{du}{\sqrt{u}} = -\sqrt{u} = -\sqrt{1 - x^{2}}.$

So now bringing in the limits, our original integral comes out to

$(\sin^{-1}(x) - \sqrt{1 - x^{2}}) \large |_{-1}^{1}$ $= \sin^{-1}(1) - \sin^{-1}(-1) = 2\sin^{-1}(1) = 2*\dfrac{\pi}{2} = \pi = \boxed{3.1415926}$

to 7 decimal places.

A slick but admittedly contrived method of solving the problem:

Note that $\large \int_{a}^{b} f(x) dx$ = $\large \int_{a}^{b} f(a+b-x) dx$ when your function is Riemann Integrable.

So $\large I = \int_{-1}^{1} \frac{\sqrt{1+x}}{\sqrt{1-x}} dx$ = $\large \int_{-1}^{1} \frac{\sqrt{1-x}}{\sqrt{1+x}}$

Then, $\large 2I = \int_{-1}^{1} \frac{\sqrt{1+x}}{\sqrt{1-x}} + \frac{\sqrt{1-x}}{\sqrt{1+x}} dx$ = $\large \int_{-1}^{1} \frac{2}{\sqrt{1-x^2}} dx$ = $\large 2(\arcsin{1} - \arcsin{-1}) = 2\pi$

Thus $\large I = \pi$

Another way to solve this integral is replacing $x$ by $\cos\theta$ , which lead us to:

$\displaystyle\int_{-1}^1 \sqrt{\frac{1+x}{1-x}}\,dx$ = $\displaystyle\int_{\pi}^0 \sqrt{\frac{1+\cos\theta}{1-\cos\theta}}(-\sin\theta)\,d\theta$ = $\displaystyle\int_{0}^\pi \frac{1}{\tan\left(\frac{\theta}{2}\right)}2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)\,d\theta$ = $2\displaystyle\int_{0}^\pi cos^2\left(\frac{\theta}{2}\right) \,d\theta$ = $2\dfrac{\pi}{2}$ = $\pi$

Note: $\tan^2\left(\dfrac{\theta}{2}\right) = \dfrac{1-\cos\theta}{1+\cos\theta}$

Another Solution:

put x= sin theta

dx becomes cos theta dtheta

The limits for theta become -pi/2 to pi/2

The integrand has sqrt(1+sin theta)/sqrt(1-sin theta).

Use the identity (sin theta/2 + cos theta/2)^2=1+sin theta and (cos theta/2-sin theta/2)^2=1-sin theta

Then multiply and divide by cos theta/2+sin theta/2.

This will change the new integrand to 1+sin theta d theta within the limits -pi/2 to pi/2

Since sin theta is an odd function, this will simplify to integral -pi/2 to pi/2 d theta=pi