Standard algebraic integrals

Some integrals show up so often, in so many disguises, that memorising them outright is faster than re-deriving them every time — the way you memorise $\frac{d}{dx}\sin x = \cos x$ rather than re-running first principles each time. This is exactly that situation for a family of integrands built from $x^2\pm a^2$ and $\sqrt{x^2\pm a^2}$ . This section belongs to the Integrals (Indefinite) section of the ISC Class 12 Calculus unit. Beyond ordinary power-rule and substitution integrals, the syllabus explicitly lists a short table of standard algebraic forms — and, crucially, the skill of recognising when a given integral is secretly one of these forms in disguise, usually after completing the square or splitting a linear numerator. That recognition step, more than the memorised formulae themselves, is what exam questions actually test.

The core insight

A handful of memorised standard forms for $\int dx/(x^2\pm a^2)$ , $\int dx/\sqrt{a^2-x^2}$ , $\int dx/\sqrt{x^2\pm a^2}$ and their square-root counterparts solve most quadratic-under-a-root integrals once a general quadratic is reshaped into $t^2\pm a^2$ by completing the square, or a linear numerator is split into a derivative-matching piece plus a constant.

The core idea

The six standard results. Each is provable by a trigonometric or hyperbolic substitution, but in practice they are used as ready-made formulae, looked up the way you would look up a derivative:

\int\frac{dx}{x^2+a^2} = \frac{1}{a}\tan^{-1}\frac{x}{a}+C, \qquad \int\frac{dx}{x^2-a^2} = \frac{1}{2a}\log\left|\frac{x-a}{x+a}\right|+C

where: $a$ is a fixed positive constant and $x$ the variable of integration; the sign in the denominator (plus or minus $a^2$ ) decides which of the two formulae applies.

\int\frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\frac{x}{a}+C, \qquad \int\frac{dx}{\sqrt{x^2+a^2}} = \log\left|x+\sqrt{x^2+a^2}\right|+C, \qquad \int\frac{dx}{\sqrt{x^2-a^2}} = \log\left|x+\sqrt{x^2-a^2}\right|+C

\int\sqrt{a^2-x^2}\,dx = \frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C, \qquad \int\sqrt{x^2\pm a^2}\,dx = \frac{x}{2}\sqrt{x^2\pm a^2}\pm\frac{a^2}{2}\log\left|x+\sqrt{x^2\pm a^2}\right|+C

Two recognition moves make the table usable on a general quadratic. First, for $ax^2+bx+c$ that is not already a perfect $t^2\pm a^2$ shape, complete the square: $ax^2+bx+c = a\left[(x+k)^2 \pm m^2\right]$ , and match the bracket against the table. Second, for a linear-over-quadratic integrand $\int\dfrac{px+q}{ax^2+bx+c}\,dx$ , split the numerator against the denominator's own derivative,

px+q = \lambda\cdot\frac{d}{dx}(ax^2+bx+c) + \mu,

where: $\lambda$ and $\mu$ are constants chosen so the equation holds identically in $x$ . This turns one integral into two: an $f'(x)/f(x)$ piece that integrates to a logarithm, and a $\mu$ -weighted standard form from the table above.

The same split extends to square roots. For $\int(px+q)\sqrt{ax^2+bx+c}\,dx$ , the identical split $px+q = \lambda\cdot\frac{d}{dx}(ax^2+bx+c)+\mu$ separates the integral into $\lambda\int(2ax+b)\sqrt{ax^2+bx+c}\,dx$ — solved by the substitution $t = ax^2+bx+c$ , collapsing to the power rule $\int\sqrt{t}\,dt$ — plus $\mu\int\sqrt{ax^2+bx+c}\,dx$ , solved by completing the square and matching one of the $\int\sqrt{x^2\pm a^2}\,dx$ forms. This "one derivative-matching piece, one standard-form piece" pattern is the single most reusable idea in this entire topic, and it resurfaces later in trigonometric-rational integrals with a linear numerator, so it is worth internalising here rather than treating it as a one-off trick.

Why memorise instead of re-derive every time. Each of the six results can, in principle, be obtained fresh with a trigonometric substitution ( $x=a\tan\theta$ , $x=a\sin\theta$ , or $x=a\sec\theta$ ) or, for the log forms, by partial fractions. But under exam time pressure, re-deriving a result from scratch for every single integral is slow and error-prone. The efficient exam strategy is therefore two-layered: know the table cold as a fast lookup, but also keep one derivation (shown below) fresh enough in memory to reconstruct any forgotten entry, and to justify a result if a question explicitly asks you to derive rather than merely evaluate.

Figure

A decision flow from a quadratic integrand to complete the square or split the numerator ending at a standard form lookup

Building the log-form result

The goal is to derive the logarithmic standard result $\int\frac{dx}{x^2-a^2}=\frac{1}{2a}\log\left|\frac{x-a}{x+a}\right|+C$ from scratch using partial fractions rather than quoting it, so that the formula's origin — and the reason the $\frac{1}{2a}$ factor sits exactly where it does — is understood well enough to reconstruct under exam pressure even if the memorised table is forgotten.

Step 1. Factor the denominator as a difference of squares: $x^2-a^2=(x-a)(x+a)$ . This is the algebraic fact that unlocks the whole derivation — a fraction with this denominator can be split into two simpler fractions, one for each factor.

Step 2. Set up the partial-fraction decomposition $\dfrac{1}{x^2-a^2}=\dfrac{A}{x-a}+\dfrac{B}{x+a}$ and clear denominators: $1 = A(x+a)+B(x-a)$ . Substituting $x=a$ gives $1=A(2a)$ , so $A=\dfrac{1}{2a}$ ; substituting $x=-a$ gives $1=B(-2a)$ , so $B=-\dfrac{1}{2a}$ .

Step 3. Integrate the two simple fractions separately: $\displaystyle\int\frac{dx}{x^2-a^2}=\frac{1}{2a}\int\frac{dx}{x-a}-\frac{1}{2a}\int\frac{dx}{x+a}=\frac{1}{2a}\log|x-a|-\frac{1}{2a}\log|x+a|+C$ .

Step 4. Combine the two logarithms using $\log m-\log n=\log(m/n)$ to reach the standard form: $\displaystyle\int\frac{dx}{x^2-a^2}=\frac{1}{2a}\log\left|\frac{x-a}{x+a}\right|+C$ . This matches the memorised table exactly, and the same steps, run in reverse, are how you would verify the result by differentiating back.

Worked example

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

Step 1. Compute the denominator's derivative: $\dfrac{d}{dx}(x^2+4x+5)=2x+4$ . Compare it with the numerator $2x+3$ to find $\lambda$ and $\mu$ such that $2x+3=\lambda(2x+4)+\mu$ . Matching the $x$ -coefficient gives $\lambda=1$ , and matching constants gives $4+\mu=3$ , so $\mu=-1$ : $2x+3=1\cdot(2x+4)-1$ .

Step 2. Split the integral using that identity:

\int\frac{2x+3}{x^2+4x+5}dx = \int\frac{2x+4}{x^2+4x+5}dx - \int\frac{dx}{x^2+4x+5}.

Step 3. The first piece is exactly the $f'(x)/f(x)$ form, which integrates directly to a logarithm: $\displaystyle\int\frac{2x+4}{x^2+4x+5}dx = \log\left|x^2+4x+5\right|+C_1$ .

Step 4. For the second piece, complete the square: $x^2+4x+5=(x+2)^2+1^2$ , matching the $\int dx/(t^2+a^2)$ standard form with $t=x+2$ , $a=1$ : $\displaystyle\int\frac{dx}{(x+2)^2+1^2} = \tan^{-1}(x+2)+C_2$ .

Step 5. Combine both pieces into the final answer: $\displaystyle\int\frac{2x+3}{x^2+4x+5}dx = \boxed{\log\left|x^2+4x+5\right| - \tan^{-1}(x+2) + C}$ .

Competitive stretch

A quick square-root check on the same table entry, with no splitting needed: $\displaystyle\int\sqrt{4-x^2}\,dx$ matches $\int\sqrt{a^2-x^2}\,dx$ directly with $a=2$ (since $4-x^2$ is already a perfect $a^2-x^2$ shape), giving $\dfrac{x}{2}\sqrt{4-x^2}+2\sin^{-1}\dfrac{x}{2}+C$ — no completing the square or splitting is needed once the integrand is already in the exact table shape.

Common traps

Mixing up the $\tan^{-1}$ and log forms — the sign inside the denominator ( $x^2+a^2$ versus $x^2-a^2$ ) decides which formula applies; check it before reaching for either.
Forgetting to complete the square on a general quadratic like $x^2+6x+13$ before applying a table formula — the formulae apply only to the exact shape $t^2\pm a^2$ .
Splitting $px+q$ incorrectly — the coefficient $\lambda$ must exactly reproduce the $x$ -term via $\lambda\cdot f'(x)$ ; get $\lambda$ wrong and the leftover constant $\mu$ is wrong too.
Dropping the $\frac{1}{a}$ or $\frac{1}{2a}$ scaling factor in front of $\tan^{-1}$ or $\log$ — these come from the chain rule and are easy to omit under time pressure.
Choosing the wrong square-root formula among the three $\int\sqrt{\cdots}\,dx$ variants — $\int\sqrt{a^2-x^2}\,dx$ gives a $\sin^{-1}$ term, while $\int\sqrt{x^2\pm a^2}\,dx$ give logarithmic terms; match the exact sign pattern under the root before substituting values of $a$ .

Ask the tutor: Want a step-by-step walkthrough of splitting a trickier linear-over-quadratic integral into its log and standard-form pieces? Ask the AI tutor for a guided example.

How do I know whether to complete the square or split the numerator first

Look at the numerator. If it is just a constant (or the integral is purely $\int dx/(ax^2+bx+c)$ or $\int dx/\sqrt{ax^2+bx+c}$ with no $x$ multiplying anything on top), go straight to completing the square. If the numerator has an $x$ term, like $\int(px+q)/(ax^2+bx+c)\,dx$ , split it first into the $f'(x)/f(x)$ piece plus a constant piece — the constant piece is then the one you complete the square on.

Why does completing the square work here but not for a cubic denominator

Completing the square converts any quadratic $ax^2+bx+c$ exactly into $a[(x+k)^2\pm m^2]$ , because a quadratic has only one turning point and its graph is always a shifted, scaled parabola — the substitution $t=x+k$ removes the linear term completely. A cubic has no such universal shifted-square form, so this trick is specific to quadratics; cubic and higher denominators generally need full partial fractions with more terms instead.

Do I need to memorise all six standard forms, or can I derive them in the exam

Memorising them is faster and safer under time pressure, but knowing the derivations protects you if you misremember one. The $\tan^{-1}$ and $\sin^{-1}$ forms come from the substitutions $x=a\tan\theta$ and $x=a\sin\theta$ respectively; the log forms for $x^2-a^2$ come from partial fractions, as shown above. If you blank on a formula, five minutes rebuilding it from these substitutions is safer than guessing.