Properties of indefinite integrals

Every integral you will ever evaluate in Class 12 begins with the same silent move: split it into pieces you already recognise. That move is not a trick — it is a small, provable set of rules called the properties of indefinite integrals, and they exist because integration is defined as the reverse of differentiation, and differentiation is linear. Sit right after the definition $\int f(x)\,dx = F(x) + C$ in the NCERT Integrals chapter, these properties tell you exactly when you may pull a constant out, break a sum apart, and merge several arbitrary constants into one. Skip them and you will try to force substitution or parts onto expressions that are really just disguised sums. Master them and the "standard list of integrals" becomes a toolbox you can combine freely, term by term, on almost every problem that follows.

The core insight

Because differentiation is linear, so is integration: an integral of a sum splits into a sum of integrals, and a constant factor slides freely across the integral sign, with only one arbitrary constant kept at the end.

The core idea

Same derivative, differ by a constant. If two functions $F(x)$ and $G(x)$ satisfy $\dfrac{d}{dx}F(x) = \dfrac{d}{dx}G(x)$ on an interval, then $F(x) - G(x) = C$ , a constant, on that interval. This is why $\int f(x)\,dx$ is written as a whole family $F(x)+C$ rather than a single function — every antiderivative of $f$ differs from every other one by a constant.

Term-by-term (sum/difference) rule. For two functions $f$ and $g$ ,

\int [f(x) \pm g(x)]\,dx = \int f(x)\,dx \pm \int g(x)\,dx

where: $f(x)$ and $g(x)$ are the pieces being combined, and each $\int$ on the right is evaluated separately before adding. The rule extends to any finite number of terms $f_1, f_2, \dots, f_n$ added or subtracted.

Constant-multiple rule. For any real constant $k$ ,

\int k\,f(x)\,dx = k\int f(x)\,dx

where: $k$ is a fixed real number — never a function of $x$ . Combining both rules gives the general linearity property used constantly in practice,

\int \left[a_1 f_1(x) + a_2 f_2(x) + \cdots + a_n f_n(x)\right] dx = a_1\!\int f_1(x)\,dx + a_2\!\int f_2(x)\,dx + \cdots + a_n\!\int f_n(x)\,dx

where: $a_1, a_2, \dots, a_n$ are real constants and $f_1, \dots, f_n$ are the functions being combined.

Only one constant survives. Each separate integral on the right technically produces its own constant of integration, but the sum of several arbitrary constants is itself just another arbitrary constant, so a single " $+C$ " is written once at the end of the whole expression — never one per term.

What these rules do not give you. There is no product rule and no quotient rule for integrals: $\int f(x)g(x)\,dx \neq \left(\int f\,dx\right)\left(\int g\,dx\right)$ , and division behaves no better. Products and quotients of non-trivial functions need dedicated techniques — substitution, partial fractions, integration by parts — covered elsewhere in the chapter; these properties only ever split sums and pull out constants.

Figure

A balance scale showing the integral of a sum splitting into two separate integrals of equal total weight

Derivation AWhy sum and constant-multiple rules hold for integrals

The goal is to show that the sum/difference and constant-multiple properties of integration are not new facts to memorise, but direct, unavoidable consequences of the corresponding sum rule and constant-multiple rule already proved for derivatives in the previous chapter — because integration is nothing but the reverse operation of differentiation.

Step 1. Recall the derivative rules. For differentiable functions $f$ and $g$ with antiderivatives $F$ and $G$ (so $F' = f$ and $G' = g$ ), differentiation already obeys $\dfrac{d}{dx}[F(x) + G(x)] = F'(x) + G'(x)$ and $\dfrac{d}{dx}[k\,F(x)] = k\,F'(x)$ for a constant $k$ . These are proved purely from the limit definition of the derivative, before any integration is discussed.

Step 2. Translate into antiderivative language. Since $F' = f$ and $G' = g$ , Step 1's first identity reads $\dfrac{d}{dx}[F(x)+G(x)] = f(x) + g(x)$ . That says $F(x)+G(x)$ is an antiderivative of $f(x)+g(x)$ — by definition, that means $\int [f(x)+g(x)]\,dx = F(x) + G(x) + C$ .

Step 3. Compare with the individual integrals. By definition $\int f(x)\,dx = F(x) + C_1$ and $\int g(x)\,dx = G(x) + C_2$ , so $\int f(x)\,dx + \int g(x)\,dx = F(x) + G(x) + (C_1+C_2)$ . Since $C_1 + C_2$ is just another arbitrary constant, this matches Step 2 exactly: $\int [f(x)+g(x)]\,dx = \int f(x)\,dx + \int g(x)\,dx$ , which is precisely the sum rule for integrals.

Step 4. Repeat for the constant multiple. Step 1's second identity gives $\dfrac{d}{dx}[k\,F(x)] = k\,f(x)$ , so $k\,F(x)$ is an antiderivative of $k\,f(x)$ , meaning $\int k\,f(x)\,dx = k\,F(x) + C = k\left(\int f(x)\,dx\right)$ after absorbing constants. Both integral properties therefore follow automatically from the linearity already established for derivatives — no separate proof is needed for the difference case either, since subtraction is addition of $-g$ , and $\int (-g)\,dx = -\int g\,dx$ by the constant-multiple rule with $k=-1$ .

Worked example

Evaluate $\displaystyle\int \left(3x^2 - 5\sin x + \frac{4}{x}\right) dx$ , then check the answer by differentiating it back.

Step 1. Split the integral term by term using the sum/difference and constant-multiple rules:

\int 3x^2\,dx - \int 5\sin x\,dx + \int \frac{4}{x}\,dx = 3\int x^2\,dx - 5\int \sin x\,dx + 4\int \frac{1}{x}\,dx

Step 2. Apply the standard results $\int x^2\,dx = \dfrac{x^3}{3}+C_1$ , $\int \sin x\,dx = -\cos x + C_2$ , and $\int \dfrac{1}{x}\,dx = \ln|x| + C_3$ to each piece:

3\cdot\frac{x^3}{3} - 5(-\cos x) + 4\ln|x| = x^3 + 5\cos x + 4\ln|x|

Step 3. Merge the three separate constants $C_1, C_2, C_3$ into a single overall constant $C$ , since their sum is itself just an arbitrary constant. Answer: $\displaystyle\int \left(3x^2 - 5\sin x + \frac{4}{x}\right) dx = x^3 + 5\cos x + 4\ln|x| + C$ .

Step 4. Check by differentiating: $\dfrac{d}{dx}\left[x^3 + 5\cos x + 4\ln|x| + C\right] = 3x^2 - 5\sin x + \dfrac{4}{x}$ , exactly the original integrand — confirming the split-and-integrate answer is correct.

Competitive stretch

Evaluate $\displaystyle\int \frac{(x+1)^2}{x}\,dx$ , where the integrand is not directly a standard form. Expand the numerator first: $\dfrac{(x+1)^2}{x} = \dfrac{x^2+2x+1}{x} = x + 2 + \dfrac{1}{x}$ . Now the properties apply directly: $\displaystyle\int \left(x+2+\frac{1}{x}\right)dx = \frac{x^2}{2} + 2x + \ln|x| + C$ . This is the real skill the topic tests — recognising when algebraic manipulation (expanding, dividing, simplifying) turns a non-standard integrand into a sum the linearity rules can finish.

Common traps

Inventing a "product rule" or "quotient rule" for integrals — no such rule exists; $\int fg\,dx \ne \int f\,dx \cdot \int g\,dx$ , and products/quotients need substitution, parts, or partial fractions instead.
Writing a separate " $+C$ " after every split term instead of one merged constant at the very end — technically each term produces one, but only a single combined $C$ should appear in the final answer.
Pulling a variable factor (something involving $x$ ) outside the integral sign as if it were a constant — the constant-multiple rule works only for a fixed real number $k$ , never for a function of $x$ .
Forgetting to algebraically simplify the integrand first (expanding a square, splitting a fraction, doing polynomial division) before hunting for a technique — many "hard-looking" integrals are just disguised sums.

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Why is there no product rule for integrals like there is for derivatives?

Differentiation has a product rule because $(fg)' = f'g + fg'$ follows directly from the limit definition applied to a product. Integration is defined as reversing differentiation, and there is no algebraic way to reverse that particular combination back into a simple product of the two separate antiderivatives. Instead, integrating a product needs a dedicated technique — usually integration by parts, which is really the product rule rearranged and integrated, or substitution when the product has a special structure.

Can I use the constant-multiple rule to pull out something like x squared from an integral, since it looks constant compared to the rest?

No. The rule $\int k f(x)\,dx = k\int f(x)\,dx$ only holds when $k$ is a genuine constant — a fixed number that does not change as $x$ varies. $x^2$ is a function of $x$ , not a constant, so pulling it outside the integral sign is not legal and will give a wrong answer. If you see a product like $x^2 \sin x$ , that calls for integration by parts or substitution, not the constant-multiple rule.

Do I need to prove the sum and constant-multiple properties every time I use them in an exam?

No. Once you have stated (or the question has assumed) that indefinite integration is linear, you can apply the sum/difference and constant-multiple rules directly without re-deriving them, exactly the way you use the derivative sum rule without re-proving it each time. The proof matters for understanding why the properties are trustworthy, and it may be asked as a standalone theory question, but routine evaluation problems expect you to use the properties, not justify them from scratch.

Properties of indefinite integrals

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads