Integral of $$$\sqrt{\frac{1 - x}{x + 1}}$$$

Find $$$\int \sqrt{\frac{1 - x}{x + 1}}\, dx$$$.

The input is rewritten: $$$\int{\sqrt{\frac{1 - x}{x + 1}} d x}=\int{\frac{\sqrt{1 - x}}{\sqrt{x + 1}} d x}$$$.

Multiply the numerator and denominator by $$$\sqrt{x + 1}$$$ and simplify:

$${\color{red}{\int{\frac{\sqrt{1 - x}}{\sqrt{x + 1}} d x}}} = {\color{red}{\int{\frac{\sqrt{1 - x^{2}}}{x + 1} d x}}}$$

Let $$$x=\sin{\left(u \right)}$$$.

Then $$$dx=\left(\sin{\left(u \right)}\right)^{\prime }du = \cos{\left(u \right)} du$$$ (steps can be seen »).

Also, it follows that $$$u=\operatorname{asin}{\left(x \right)}$$$.

Integrand becomes

$$$\frac{\sqrt{1 - x^{2}}}{x + 1} = \frac{\sqrt{1 - \sin^{2}{\left( u \right)}}}{\sin{\left( u \right)} + 1}$$$

Use the identity $$$1 - \sin^{2}{\left( u \right)} = \cos^{2}{\left( u \right)}$$$:

$$$\frac{\sqrt{1 - \sin^{2}{\left( u \right)}}}{\sin{\left( u \right)} + 1}=\frac{\sqrt{\cos^{2}{\left( u \right)}}}{\sin{\left( u \right)} + 1}$$$

Assuming that $$$\cos{\left( u \right)} \ge 0$$$, we obtain the following:

$$$\frac{\sqrt{\cos^{2}{\left( u \right)}}}{\sin{\left( u \right)} + 1} = \frac{\cos{\left( u \right)}}{\sin{\left( u \right)} + 1}$$$

Therefore,

$${\color{red}{\int{\frac{\sqrt{1 - x^{2}}}{x + 1} d x}}} = {\color{red}{\int{\frac{\cos^{2}{\left(u \right)}}{\sin{\left(u \right)} + 1} d u}}}$$

Rewrite the cosine in terms of the sine, rewrite the numerator further, use the formula for difference of squares, and simplify:

$${\color{red}{\int{\frac{\cos^{2}{\left(u \right)}}{\sin{\left(u \right)} + 1} d u}}} = {\color{red}{\int{\left(1 - \sin{\left(u \right)}\right)d u}}}$$

Integrate term by term:

$${\color{red}{\int{\left(1 - \sin{\left(u \right)}\right)d u}}} = {\color{red}{\left(\int{1 d u} - \int{\sin{\left(u \right)} d u}\right)}}$$

Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=1$$$:

$$- \int{\sin{\left(u \right)} d u} + {\color{red}{\int{1 d u}}} = - \int{\sin{\left(u \right)} d u} + {\color{red}{u}}$$

The integral of the sine is $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:

$$u - {\color{red}{\int{\sin{\left(u \right)} d u}}} = u - {\color{red}{\left(- \cos{\left(u \right)}\right)}}$$

Recall that $$$u=\operatorname{asin}{\left(x \right)}$$$:

$$\cos{\left({\color{red}{u}} \right)} + {\color{red}{u}} = \cos{\left({\color{red}{\operatorname{asin}{\left(x \right)}}} \right)} + {\color{red}{\operatorname{asin}{\left(x \right)}}}$$

Therefore,

$$\int{\frac{\sqrt{1 - x}}{\sqrt{x + 1}} d x} = \sqrt{1 - x^{2}} + \operatorname{asin}{\left(x \right)}$$

Add the constant of integration:

$$\int{\frac{\sqrt{1 - x}}{\sqrt{x + 1}} d x} = \sqrt{1 - x^{2}} + \operatorname{asin}{\left(x \right)}+C$$

$$$\int \sqrt{\frac{1 - x}{x + 1}}\, dx = \left(\sqrt{1 - x^{2}} + \operatorname{asin}{\left(x \right)}\right) + C$$$A