Evaluate the integral of ((a+x)/(a-x))1/2

Evaluate the following integral for $a > 0$ .

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

First, we multiply the numerator and denominator in the fraction inside the square root by $a+x$ and do some simplification,

$\begin{align*} \int \sqrt{\frac{a+x}{a-x}} \, dx &= \int \sqrt{\frac{(a+x)^2}{(a-x)(a+x)}} \, dx \\[9pt] &= \int \frac{a+x}{\sqrt{a^2-x^2}} \, dx \\[9pt] &= \int \frac{a}{\sqrt{a^2-x^2}} \,dx + \int \frac{x}{\sqrt{a^2-x^2}} \, dx \\[9pt] &= \int \frac{1}{\sqrt{1 - \left( \frac{x}{a} \right)^2}} \, dx + \int \frac{x}{\sqrt{a^2-x^2}} \, dx. \end{align*}$

Now, we evaluate these two integrals separately. For the first integral we make the substitution $u = \frac{x}{a}$ which implies $du = \frac{1}{a} \, dx$ . Therefore,

$\begin{align*} \int \frac{1}{\sqrt{1 - \left( \frac{x}{a} \right)^2}} \, dx &= a \int \frac{1}{\sqrt{1-u^2}} \, du \\ &= a \arcsin u + C \\ &= a \arcsin \left( \frac{x}{a} \right) + C. \end{align*}$

For the second integral, we make the substitution $u = a^2 - x^2$ which implies $du = -2x \, dx$ . Therefore,

$\begin{align*} \int \frac{x}{\sqrt{a^2 - x^2}} \, dx &= -\frac{1}{2} \int \frac{1}{\sqrt{u}} \, du \\[9pt] &= - \sqrt{u} + C \\ &= -\sqrt{a^2 - x^2} + C. \end{align*}$

Putting these together we then have,

$\begin{align*} \int \sqrt{\frac{a+x}{a-x}} \, dx &= \int \frac{1}{\sqrt{1- \left( \frac{x}{a} \right)^2}} \, dx + \int \frac{x}{\sqrt{a^2 - x^2}} \, dx \\ &= a \arcsin \left( \frac{x}{a} \right) - \sqrt{a^2-x^2} + C. \end{align*}$

Stumbling Robot

Evaluate the integral of ((a+x)/(a-x))^1/2

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