Q.Evaluate: $\int \sqrt{\dfrac{a+x}{a-x}}\,dx$

Concept: Integration By Completing Square (here, we rationalise the integrand by multiplying numerator and denominator, then use a trigonometric substitution).

Step 1: Multiply numerator and denominator inside the square root by $a+x$:

Step 2: Split the integral:

$\int \frac{a+x}{\sqrt{a^2 - x^2}}\,dx = a\int\frac{dx}{\sqrt{a^2 - x^2}} + \int\frac{x\,dx}{\sqrt{a^2 - x^2}}.$ …

The key insight here is that the integrand $\sqrt{\frac{a+x}{a-x}}$ looks like it might be related to the derivative of an inverse trigonometric function, but it's not in a standard form. The trick is to rationalize the denominator by multiplying the numerator and denominator inside the square root by $\sqrt{a+x}$. This transforms the expression into something that involves $\sqrt{a^2 - x^2}$, which is a classic signpost for a trigonometric substitution or a direct standard integral.

Let's work through it step by step.

1. Rewrite the integrand algebraically. Multiply the numerator and denominator inside the square root by $\sqrt{a+x}$:

This is valid for $|x| < a$, which is the domain where the original square root is real. The expression is now a sum of two simpler terms.

2. Split the integral into two manageable parts.

The integral becomes:

Now we have two standard forms. The first is a direct inverse sine integral. The second is a simple substitution.

3. Solve the first integral.

The standard result is:

So the first term contributes $a \sin^{-1}\left(\frac{x}{a}\right)$.

4. Solve the second integral using substitution.

Let $u = a^2 - x^2$. Then $du = -2x\,dx$, so $x\,dx = -\frac{1}{2} du$.

The integral becomes: …

Method: Rationalising $\sqrt{\frac{a+x}{a-x}}$ by multiplying inside the root

Use this for integrands of the form $\sqrt{\dfrac{a+x}{a-x}}$ (or its reciprocal). Multiply numerator and denominator inside the root by the same factor to create a standard $\sqrt{a^2-x^2}$ denominator.

Steps

Step 1: Multiply top and bottom inside the root by $(a+x)$.

Step 2: Split into two standard integrals.

Step 3: Integrate each with its standard form. …

Common Mistakes

Mistake 1: Not rationalising the root.

Why it's wrong: $\sqrt{\tfrac{a+x}{a-x}}$ is hard to integrate directly; multiplying by $\tfrac{a+x}{a+x}$ inside converts it to $\tfrac{a+x}{\sqrt{a^2-x^2}}$. Correct approach: rationalise first.

Mistake 2: Wrong sign on $\int\tfrac{x}{\sqrt{a^2-x^2}}\,dx$.

Why it's wrong: this integral is $-\sqrt{a^2-x^2}$ (from $u=a^2-x^2$, $du=-2x\,dx$); a lost minus flips the term. Correct approach: track the $-\tfrac12$ from $du$. …