Q.Find the following integrals:

Concept: Standard forms – these are direct applications of the inverse hyperbolic / logarithmic and inverse trigonometric integration formulas.

(i) $\int \dfrac{dx}{x^2 - 16}$

Factor the denominator: $x^2 - 16 = (x-4)(x+4)$. Use partial fractions or the standard formula

$\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \log\left|\frac{x-a}{x+a}\right| + C$.

Here $a = 4$, so:

(ii) $\int \dfrac{dx}{\sqrt{2x - x^2}}$

Complete the square: $2x - x^2 = -(x^2 - 2x) = -(x-1)^2 + 1 = 1 - (x-1)^2$.

Thus the integral becomes $\int \frac{dx}{\sqrt{1 - (x-1)^2}}$, which matches $\int \frac{du}{\sqrt{a^2 - u^2}} = \sin^{-1}\frac{u}{a} + C$ with $a=1$, $u = x-1$.

The key to both problems is recognising that they match the standard forms from your integral tables — but with a slight twist. For (i), the denominator is a difference of squares, which screams for partial fractions or the inverse hyperbolic tangent formula. For (ii), the expression under the square root is a quadratic that doesn’t immediately look like $1 - u^2$, but completing the square will make it so.

Let’s work through each one.

(i) $\int \dfrac{dx}{x^2 - 16}$

1. Recognise the standard form.

You know that $\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \log\left|\frac{x-a}{x+a}\right| + C$. Here $a^2 = 16$, so $a = 4$. This is a direct match — no substitution needed.

2. Apply the formula.

Plug $a = 4$ into the formula:

3. Done.

No further simplification is needed. The absolute value ensures the logarithm is defined for $x$ outside the interval $(-4, 4)$ as well.

(ii) $\int \dfrac{dx}{\sqrt{2x - x^2}}$

1. Complete the square inside the square root.

The expression $2x - x^2$ is a quadratic. Write it as:

So the integral becomes:

2. Recognise the standard form.

Now it matches $\int \frac{du}{\sqrt{1 - u^2}} = \sin^{-1} u + C$, with $u = x-1$ and $du = dx$.

3. Substitute and integrate.

Let $u = x-1$, then $du = dx$. The integral is:

4. Done.

No constant factor appears because the completed square gave exactly $1 - (x-1)^2$ — the coefficient of $u^2$ is $1$.

Method: Complete the Square, Then Match a Standard Form

Use this for integrals of $\dfrac{1}{x^2\pm a^2}$, $\dfrac{1}{\sqrt{a^2-x^2}}$ type expressions, especially when the quadratic has a linear term hidden in it.

Steps

Step 1: Recognise or create a pure quadratic-in-$(x-h)$.

If the denominator is $x^2-a^2$, it is already a difference of squares. If it is $2x-x^2$ or similar, complete the square first: $2x-x^2=1-(x-1)^2$.

Step 2: Apply the matching standard formula.

Use the appropriate table result:

Step 3: Substitute $u=x-h$ if you completed the square, then finish.

For $\dfrac{1}{\sqrt{2x-x^2}}=\dfrac{1}{\sqrt{1-(x-1)^2}}$, use $u=x-1$ to get $\sin^{-1}(x-1)+C$.

Common Mistakes

Mistake 1: Confusing $\dfrac{1}{x^2-a^2}$ with $\dfrac{1}{x^2+a^2}$.

Why it's wrong: the minus form gives a logarithm, the plus form gives $\frac1a\tan^{-1}\frac{x}{a}$. Correct approach: check the sign before choosing the formula.

Mistake 2: Not completing the square for $2x-x^2$.

Why it's wrong: it does not match $\sqrt{a^2-u^2}$ until rewritten as $1-(x-1)^2$. Correct approach: complete the square to expose the standard form.

Mistake 3: Getting the $\dfrac{1}{2a}$ factor wrong in the log formula.

Why it's wrong: with $a=4$ the factor is $\frac{1}{8}$; omitting it or using $\frac14$ scales the answer wrongly. Correct approach: use $\frac{1}{2a}$ exactly.