Integrals of Some Particular Functions

The Six Standard Special Forms

Six integrals appear so often that NCERT promotes them to standard formulae — used from now on without proof:

1. $\displaystyle\int \frac{dx}{x^2 - a^2} = \frac{1}{2a}\log\left|\frac{x - a}{x + a}\right| + C$
2. $\displaystyle\int \frac{dx}{a^2 - x^2} = \frac{1}{2a}\log\left|\frac{a + x}{a - x}\right| + C$
3. $\displaystyle\int \frac{dx}{x^2 + a^2} = \frac{1}{a}\tan^{-1}\frac{x}{a} + C$
4. $\displaystyle\int \frac{dx}{\sqrt{x^2 - a^2}} = \log\left|x + \sqrt{x^2 - a^2}\right| + C$
5. $\displaystyle\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac{x}{a} + C$
6. $\displaystyle\int \frac{dx}{\sqrt{x^2 + a^2}} = \log\left|x + \sqrt{x^2 + a^2}\right| + C$

Where they come from: (1) and (2) split by partial fractions — e.g. $\frac{1}{x^2 - a^2} = \frac{1}{2a}\left[\frac{1}{x-a} - \frac{1}{x+a}\right]$; (3) uses the substitution $x = a\tan\theta$; (4) uses $x = a\sec\theta$; (5) uses $x = a\sin\theta$; (6) uses $x = a\tan\theta$ — each trig substitution collapses the denominator via a Pythagorean identity.

How to tell (1) from (2) and the log forms apart: look at what can vanish. $x^2 - a^2$ (roots at $\pm a$, log of a ratio with $x - a$ on top); $a^2 - x^2$ (same roots, $a + x$ on top); $x^2 + a^2$ (never zero — that's the $\tan^{-1}$). Under a square root: minus in front of $x^2$ → $\sin^{-1}$; plus → a log.

Key Point: Match the SIGN PATTERN, then read off the answer. $x^2 - a^2$, $a^2 - x^2$ and $x^2 + a^2$ each have exactly one formula, with and without the square root.

[JEE Tip] Formulas (4) and (6) are often written with $\sinh^{-1}, \cosh^{-1}$ in other books; JEE answer keys use the log forms — quote them exactly as above.

General Quadratics: Complete the Square

Any integral $\displaystyle\int\frac{dx}{ax^2 + bx + c}$ or $\displaystyle\int\frac{dx}{\sqrt{ax^2 + bx + c}}$ reduces to one of the six standard forms by completing the square:

$$ax^2 + bx + c = a\left[\left(x + \frac{b}{2a}\right)^2 + \left(\frac{c}{a} - \frac{b^2}{4a^2}\right)\right]$$

Put $t = x + \frac{b}{2a}$ and write the leftover constant as $\pm k^2$; the integral becomes $\frac{1}{a}\int\frac{dt}{t^2 \pm k^2}$ (or its square-root version) — a table entry.

Worked micro-example: $x^2 - 6x + 13 = (x-3)^2 + 4$, so $\int\frac{dx}{x^2 - 6x + 13} = \frac{1}{2}\tan^{-1}\frac{x-3}{2} + C$.

Linear-over-Quadratic: the A-B Split

For $\displaystyle\int\frac{px + q}{ax^2 + bx + c}\,dx$ (and the same with a square root), write the numerator as

$$px + q = A\cdot\frac{d}{dx}(ax^2 + bx + c) + B = A(2ax + b) + B$$

and find $A, B$ by comparing coefficients. The integral splits into two solvable pieces:

• the $A$-piece is $\int\frac{(\text{derivative of denominator})}{\text{denominator}}dx = \log|\text{denominator}|$ (or $2\sqrt{\text{denominator}}$ under a root),
• the $B$-piece is a pure standard form after completing the square.

Key Point: Two moves cover this entire section: complete the square (no $x$ in the numerator) and numerator = A(derivative) + B (linear numerator). Identify which case you're in before touching the algebra.

[Board Important] When completing the square with a negative $x^2$ coefficient — e.g. $5 - 4x - x^2 = 9 - (x+2)^2$ — factor the minus sign carefully first. Sign slips here are the top mark-loser of Exercise 7.4.

Solved Examples

Example 1: Straight table match

Find $\displaystyle\int \frac{dx}{x^2 - 16}$.

Solution:

1. Match the pattern: $x^2 - a^2$ with $a = 4$ — standard form (1).
2. Apply: $\frac{1}{2(4)}\log\left|\frac{x-4}{x+4}\right| + C$.

Final Answer: $\frac{1}{8}\log\left|\frac{x - 4}{x + 4}\right| + C$.

Example 2: Complete the square under a root

Find $\displaystyle\int \frac{dx}{\sqrt{2x - x^2}}$.

Solution:

1. Complete the square: $2x - x^2 = 1 - (x - 1)^2$.
2. Substitute: $t = x - 1$: the integral is $\int\frac{dt}{\sqrt{1 - t^2}} = \sin^{-1}t + C$.
3. Return: $\sin^{-1}(x - 1) + C$.

Final Answer: $\sin^{-1}(x - 1) + C$.

Takeaway: Minus sign on $x^2$ under a root → expect $\sin^{-1}$. The completed square tells you the shifted variable and the radius.

Example 3: Quadratic that never vanishes

Find $\displaystyle\int \frac{dx}{x^2 - 6x + 13}$.

Solution:

1. Complete the square: $x^2 - 6x + 13 = (x - 3)^2 + 2^2$.
2. Substitute: $t = x - 3$: $\int\frac{dt}{t^2 + 2^2} = \frac{1}{2}\tan^{-1}\frac{t}{2} + C$.
3. Return: $\frac{1}{2}\tan^{-1}\frac{x - 3}{2} + C$.

Final Answer: $\frac{1}{2}\tan^{-1}\left(\frac{x-3}{2}\right) + C$.

Takeaway: Positive leftover constant ($+4$) → the $\tan^{-1}$ form. Negative → the log-ratio form (next example).

Example 4: Factorisable quadratic

Find $\displaystyle\int \frac{dx}{3x^2 + 13x - 10}$.

Solution:

1. Take out the 3 and complete the square: $3x^2 + 13x - 10 = 3\left[\left(x + \frac{13}{6}\right)^2 - \left(\frac{17}{6}\right)^2\right]$.
2. Apply form (1) with $t = x + \frac{13}{6}$, $k = \frac{17}{6}$: $\frac{1}{3}\cdot\frac{1}{2k}\log\left|\frac{t - k}{t + k}\right| = \frac{1}{17}\log\left|\frac{x - \frac{4}{6}\cdot\frac{6}{6}}{\ldots}\right|$ — simplifying the ratio: $\frac{6x - 4}{6x + 30}$.
3. Simplify and absorb constants: $\frac{1}{17}\log\left|\frac{3x - 2}{x + 5}\right| + C$.

Final Answer: $\frac{1}{17}\log\left|\frac{3x - 2}{x + 5}\right| + C$.

Takeaway: The constant $\frac{1}{17}\log\frac{1}{3}$ produced while cancelling the 6's merges into $C$ — tidy ratios are always legal up to a constant.

Example 5: Root of a quadratic with no constant term

Find $\displaystyle\int \frac{dx}{\sqrt{5x^2 - 2x}}$.

Solution:

1. Factor out 5: $\sqrt{5x^2 - 2x} = \sqrt{5}\sqrt{x^2 - \frac{2x}{5}}$.
2. Complete the square: $x^2 - \frac{2x}{5} = \left(x - \frac{1}{5}\right)^2 - \left(\frac{1}{5}\right)^2$.
3. Apply form (4): $\frac{1}{\sqrt 5}\log\left|t + \sqrt{t^2 - \frac{1}{25}}\right|$ with $t = x - \frac{1}{5}$.

Final Answer: $\frac{1}{\sqrt{5}}\log\left|x - \frac{1}{5} + \sqrt{x^2 - \frac{2x}{5}}\right| + C$.

Takeaway: The $\frac{1}{\sqrt a}$ factor from pulling $a$ out of the root is easy to forget — write the factoring step explicitly.

Example 6: Linear numerator over a quadratic

Find $\displaystyle\int \frac{x + 2}{2x^2 + 6x + 5}\,dx$.

Solution:

1. Split the numerator: $x + 2 = A(4x + 6) + B$ (using the derivative of the denominator). Comparing: $4A = 1$, $6A + B = 2$, so $A = \frac{1}{4}$, $B = \frac{1}{2}$.
2. A-piece: $\frac{1}{4}\int\frac{4x + 6}{2x^2 + 6x + 5}dx = \frac{1}{4}\log|2x^2 + 6x + 5|$.
3. B-piece: $\frac{1}{2}\int\frac{dx}{2x^2 + 6x + 5} = \frac{1}{4}\int\frac{dx}{\left(x + \frac{3}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \frac{1}{2}\tan^{-1}(2x + 3)$.
4. Combine: add the two pieces.

Final Answer: $\frac{1}{4}\log|2x^2 + 6x + 5| + \frac{1}{2}\tan^{-1}(2x + 3) + C$.

Takeaway: The split is deliberately built so the $A$-piece is a pure log and the $B$-piece a pure standard form — never integrate a linear-over-quadratic in one go.

Example 7: Linear numerator under a root

Find $\displaystyle\int \frac{x + 3}{\sqrt{5 - 4x - x^2}}\,dx$.

Solution:

1. Split: $x + 3 = A(-4 - 2x) + B$; comparing: $-2A = 1$, $-4A + B = 3$, so $A = -\frac{1}{2}$, $B = 1$.
2. A-piece: $-\frac{1}{2}\int\frac{-4 - 2x}{\sqrt{5 - 4x - x^2}}dx = -\frac{1}{2}\cdot 2\sqrt{5 - 4x - x^2} = -\sqrt{5 - 4x - x^2}$ (since $\int \frac{dt}{\sqrt t} = 2\sqrt t$).
3. B-piece: $\int\frac{dx}{\sqrt{9 - (x+2)^2}} = \sin^{-1}\frac{x + 2}{3}$.

Final Answer: $-\sqrt{5 - 4x - x^2} + \sin^{-1}\left(\frac{x + 2}{3}\right) + C$.

Takeaway: Under a root, the $A$-piece integrates to $2\sqrt{\text{denominator}}$, not a log — the only difference from Example 6's template.

Example 8: A shifted-square root form

Find $\displaystyle\int \frac{dx}{\sqrt{(2 - x)^2 + 1}}$.

Solution:

1. Substitute the shift: $t = x - 2$ (note $(2-x)^2 = (x-2)^2$), $dt = dx$.
2. Apply form (6): $\int\frac{dt}{\sqrt{t^2 + 1}} = \log\left|t + \sqrt{t^2 + 1}\right| + C$.
3. Return: $\log\left|x - 2 + \sqrt{x^2 - 4x + 5}\right| + C$.

Final Answer: $\log\left|x - 2 + \sqrt{x^2 - 4x + 5}\right| + C$.

Example 9: A trig-disguised standard form

Find $\displaystyle\int \frac{\sec^2 x}{\sqrt{\tan^2 x + 4}}\,dx$.

Solution:

1. Spot the pair: $t = \tan x$ gives $dt = \sec^2 x\,dx$ — the numerator exactly.
2. Apply form (6): $\int\frac{dt}{\sqrt{t^2 + 2^2}} = \log\left|t + \sqrt{t^2 + 4}\right| + C$.
3. Return: $\log\left|\tan x + \sqrt{\tan^2 x + 4}\right| + C$.

Final Answer: $\log\left|\tan x + \sqrt{\tan^2 x + 4}\right| + C$.

Takeaway: Substitution and the six forms compose: first collapse the trig layer, then table-match.

Example 10: Product of two linear factors under a root

Find $\displaystyle\int \frac{dx}{\sqrt{(x - 1)(x - 2)}}$.

Solution:

1. Expand: $(x-1)(x-2) = x^2 - 3x + 2$.
2. Complete the square: $x^2 - 3x + 2 = \left(x - \frac{3}{2}\right)^2 - \left(\frac{1}{2}\right)^2$.
3. Apply form (4): $\log\left|x - \frac{3}{2} + \sqrt{x^2 - 3x + 2}\right| + C$.

Final Answer: $\log\left|x - \frac{3}{2} + \sqrt{x^2 - 3x + 2}\right| + C$.

Example 11: Full A-B split with $\sin^{-1}$ finish

Find $\displaystyle\int \frac{x + 2}{\sqrt{4x - x^2}}\,dx$.

Solution:

1. Split: $x + 2 = A(4 - 2x) + B$: $-2A = 1$, $4A + B = 2$ give $A = -\frac{1}{2}$, $B = 4$.
2. A-piece: $-\frac{1}{2}\int\frac{4 - 2x}{\sqrt{4x - x^2}}dx = -\frac{1}{2}\cdot 2\sqrt{4x - x^2} = -\sqrt{4x - x^2}$.
3. B-piece: $4\int\frac{dx}{\sqrt{4 - (x - 2)^2}} = 4\sin^{-1}\frac{x - 2}{2}$.

Final Answer: $-\sqrt{4x - x^2} + 4\sin^{-1}\left(\frac{x - 2}{2}\right) + C$.

Takeaway: Every Exercise 7.4 linear-over-root problem is this same two-piece machine — only the numbers change.