Integrals of additional types

Some integrals refuse to fall under the basic standard-function list — a fraction with $x^2+a^2$ sitting in the denominator, a square root of $a^2-x^2$ , or a general quadratic $ax^2+bx+c$ tucked under a root. CBSE Class 12 collects exactly these recurring shapes into a short, syllabus-listed set called integrals of additional types, each with a memorized closed-form result. The trick is not deriving these from scratch every time — it is recognizing the disguise. A quadratic denominator, once you complete the square, always collapses into one of a handful of boxed forms; a linear numerator, once split into "derivative of denominator plus a constant," reduces to a logarithm plus one of those same boxed forms. This section walks through the full boxed list, the completing-the-square strategy that unlocks every quadratic-in-disguise problem, and a fully worked pair of examples showing the constant-numerator and linear-numerator cases side by side.

The core insight

Every "additional type" integral is one of six boxed standard forms in $x^2\pm a^2$ or $\sqrt{a^2\pm x^2}$ , reached by completing the square on a general quadratic and, for linear numerators, splitting off the derivative of the denominator first.

The core idea

The boxed list. CBSE lists six standard results that do not come from the basic integration-of-standard-functions table. Two are reciprocal-quadratic forms:

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

where: $a$ is a positive constant fixed by the problem, and $C$ is the constant of integration. Two more involve a square root in the denominator:

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

and one is the arcsine form used constantly in area problems:

\int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\frac{x}{a}+C

Finally, the two "area under a root" formulas integrate the square root itself rather than its reciprocal:

\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-a^2}\,dx = \frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\ln\left|x+\sqrt{x^2-a^2}\right|+C

Spotting the disguise. A general quadratic $ax^2+bx+c$ rarely shows up as $x^2\pm a^2$ directly — it needs completing the square first: factor out the leading coefficient from the $x^2$ and $x$ terms, then write $ax^2+bx+c = a\left[(x+k)^2\pm m^2\right]$ for suitable constants $k$ and $m$ . Substituting $t = x+k$ turns the integral into one of the six boxed forms above, now in the variable $t$ .

Linear numerators. When the integrand is $\dfrac{px+q}{ax^2+bx+c}$ rather than a constant over the quadratic, the numerator is rewritten as a multiple of the denominator's derivative plus a leftover constant:

px+q = A(2ax+b) + B

Comparing coefficients of $x$ and the constant term on both sides pins down $A$ and $B$ . The $A(2ax+b)$ piece integrates directly by the substitution $u = ax^2+bx+c$ (giving a logarithm), and the leftover $B$ piece is a constant times one of the boxed reciprocal-quadratic forms — handled exactly as in the completing-the-square case above.

Figure

Two integrand shapes and the boxed forms they reduce to after completing the square or splitting the numerator

Why it works

Every boxed formula in this list is provable by a single trigonometric or hyperbolic substitution, and seeing the pattern once explains why the results pair up the way they do. For $\displaystyle\int \frac{dx}{\sqrt{a^2-x^2}}$ , set $x = a\sin\theta$ , so $dx = a\cos\theta\,d\theta$ and $\sqrt{a^2-x^2} = \sqrt{a^2-a^2\sin^2\theta} = a\cos\theta$ (using $\sin^2\theta+\cos^2\theta=1$ ). The integral becomes $\displaystyle\int \frac{a\cos\theta\,d\theta}{a\cos\theta} = \int d\theta = \theta + C$ , and since $\theta = \sin^{-1}(x/a)$ , the result follows immediately. The other five formulas come from the tangent substitution $x=a\tan\theta$ (for the $x^2+a^2$ family) or the secant/hyperbolic-style substitution (for the $x^2-a^2$ family) — the sine substitution matches $a^2-x^2$ because sine is bounded between $-1$ and $1$ , exactly matching the constraint $x\le a$ built into that square root, while tangent and secant range over all reals or beyond $1$ , matching the unbounded $x^2\pm a^2$ forms. CBSE does not require re-deriving these formulas on every use — they are meant to be memorized and applied directly — but knowing this substitution machinery underneath explains why arcsine pairs with $a^2-x^2$ , arctangent with $x^2+a^2$ , and the logarithm with $x^2-a^2$ , rather than the pairings being arbitrary.

Derivation ACompleting the square to reach a boxed form

Show that the general quadratic-denominator integral $\int \dfrac{dx}{x^2+4x+13}$ reduces, after completing the square on the denominator and substituting the shifted variable, to the boxed arctangent form $\int \dfrac{dt}{t^2+a^2} = \dfrac{1}{a}\tan^{-1}\dfrac{t}{a}+C$ , and read off the final answer in terms of $x$ .

Step 1. Complete the square on the denominator. Take the coefficient of $x$ , which is $4$ , halve it to get $2$ , and square that to get $4$ : $x^2+4x+13 = (x^2+4x+4) + 9 = (x+2)^2+9$ .

Step 2. Write $9 = 3^2$ so the denominator is manifestly of the form (shifted variable) $^2 + a^2$ : $(x+2)^2+3^2$ , giving the integral $\displaystyle\int \frac{dx}{(x+2)^2+3^2}$ .

Step 3. Substitute $t = x+2$ , so $dt = dx$ since the shift is by a constant. The integral becomes $\displaystyle\int \frac{dt}{t^2+3^2}$ , which is exactly the boxed form $\int \dfrac{dt}{t^2+a^2}$ with $a=3$ .

Step 4. Apply the boxed result directly: $\displaystyle\int \frac{dt}{t^2+3^2} = \frac{1}{3}\tan^{-1}\frac{t}{3}+C$ .

Step 5. Substitute back $t=x+2$ to express the answer in the original variable: $\displaystyle\int \frac{dx}{x^2+4x+13} = \frac13\tan^{-1}\left(\frac{x+2}{3}\right)+C$ .

Worked example

Evaluate $\displaystyle\int \frac{2x+3}{x^2+4x+13}\,dx$ , reusing the completed-square denominator from the derivation above.

Step 1. The denominator is the same as before, $x^2+4x+13$ , whose derivative is $\dfrac{d}{dx}(x^2+4x+13) = 2x+4$ . Since the numerator here is linear, write it as a multiple of this derivative plus a constant: $2x+3 = A(2x+4) + B$ .

Step 2. Compare coefficients of $x$ : matching the $2x$ terms gives $A=1$ . Compare constant terms: $4A+B = 3$ , so $4(1)+B=3$ , giving $B=-1$ . Hence $2x+3 = 1\cdot(2x+4) - 1$ .

Step 3. Split the integral using this decomposition: $\displaystyle\int \frac{2x+3}{x^2+4x+13}\,dx = \int \frac{2x+4}{x^2+4x+13}\,dx - \int \frac{dx}{x^2+4x+13}$ .

Step 4. The first integral is a direct substitution $u=x^2+4x+13$ , $du=(2x+4)\,dx$ , giving $\displaystyle\int \frac{2x+4}{x^2+4x+13}\,dx = \ln|x^2+4x+13|+C_1$ .

Step 5. The second integral is precisely the result already found in the derivation above: $\dfrac13\tan^{-1}\left(\dfrac{x+2}{3}\right)+C_2$ . Combining both pieces gives the final answer:

\int \frac{2x+3}{x^2+4x+13}\,dx = \boxed{\ln|x^2+4x+13| - \frac13\tan^{-1}\left(\frac{x+2}{3}\right) + C}

Competitive stretch

The same completed-square denominator $(x+2)^2+3^2$ also unlocks $\displaystyle\int \sqrt{x^2+4x+13}\,dx$ : substitute $t=x+2$ to get $\displaystyle\int \sqrt{t^2+3^2}\,dt$ , a shape not in the boxed list above but reachable by the analogous area-formula pattern (the $x^2+a^2$ counterpart of $\int\sqrt{x^2-a^2}\,dx$ ), reinforcing that one completed square often opens several related integrals at once.

Common traps

Confusing $\int\frac{dx}{x^2-a^2}$ (a logarithm of a ratio) with $\int\frac{dx}{x^2+a^2}$ (an arctangent) — the two look almost identical but are structurally different results.
Making a sign or arithmetic slip while completing the square, which silently sends the problem to the wrong " $-a^2$ " or " $+a^2$ " family and produces a plausible but wrong final formula.
Trying to force a linear-numerator integral $\int\frac{px+q}{ax^2+bx+c}dx$ directly onto a boxed formula without first splitting off the derivative-of-denominator piece — the boxed forms only work for constant numerators.
Leaving the leading coefficient $a$ outside the bracket incorrectly when $a\neq1$ in $ax^2+bx+c$ — it must be factored out of both the $x^2$ and $x$ terms before completing the square inside the bracket.

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Why do we need six different boxed formulas instead of one general rule?

Each formula matches a different sign pattern under the root or in the denominator — $x^2-a^2$ , $x^2+a^2$ , and $a^2-x^2$ behave differently because subtracting versus adding $a^2$ changes whether the expression can hit zero, go negative, or stay bounded. These sign differences are exactly why one substitution (sine, tangent, or a hyperbolic-style one) works for one family but not another, so no single closed form covers all three cases at once.

Do I need to memorize all six boxed formulas for exams?

Yes — CBSE expects these as recall, not re-derivation, in the Integrals chapter. What matters more than rote memorization, though, is fast pattern recognition: given a messy integral, quickly telling whether completing the square will land you on $x^2-a^2$ , $x^2+a^2$ , or $a^2-x^2$ is what actually decides whether you finish the problem in time.

What if the leading coefficient a in ax squared plus bx plus c is negative?

Factor out the negative sign along with $a$ before completing the square — this often flips which boxed family you land in, for instance turning an $x^2-a^2$ -type denominator into an $a^2-x^2$ -type one. Track the sign carefully through the factoring step, since a dropped negative sign is one of the most common ways this topic loses easy marks.