∫(x + 2)dx = (a) (x + 2)³ + k (b) (x²)/(2) + k (c) – Class 12 Mathematics | ConceptFirst

Question of 373

\int(x + 2)\,dx =

(a)

(x + 2)^3 + k

(b)

\frac{x^2}{2} + k

(c)

\frac{x^2}{2} + 2x + k

(d)

\log(x + 2) + k

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Concept understanding — Antiderivative Of Sum

The Intuition: "Differentiation distributes, so integration should too"

Suppose your speed has two parts: you speed up from excitement (part A) and slow from tiredness (part B). Your total speed is the sum. Your total distance — the antiderivative of speed — is then the distance from part A plus the distance from part B. That's the core idea: the antiderivative of a sum is the sum of the antiderivatives.

Note

This works because differentiation is linear: $\frac{d}{dx}[f(x)+g(x)] = f'(x)+g'(x)$ . Integration reverses it, so it inherits the linearity.

The Precise Statement

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

The indefinite integral of a sum of two functions equals the sum of their individual antiderivatives. This holds for any $f$ and $g$ that have antiderivatives. The same rule applies to subtraction:

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

Why It's True (A Quick Proof)

Let $F'(x) = f(x)$ and $G'(x) = g(x)$ . Consider $H(x) = F(x) + G(x)$ :

H'(x) = F'(x) + G'(x) = f(x) + g(x)

So $H(x)$ is an antiderivative of $f(x)+g(x)$ — exactly the statement.

Watch out

Each separate antiderivative has its own constant, but two constants combine into one, so we write:

$\int [f(x)+g(x)]\,dx = F(x)+G(x)+C$

A Concrete Example

Find $\int (x^2 + \cos x) \, dx$ .

Step 1: Apply the sum rule: $\int x^2 \, dx + \int \cos x \, dx$

Step 2: Each antiderivative: $\int x^2 \, dx = \frac{x^3}{3}$ , $\int \cos x \, dx = \sin x$

Step 3: Combine:

\int (x^2 + \cos x) \, dx = \frac{x^3}{3} + \sin x + C

Tip

In practice you never write separate constants — find each antiderivative and add a single $+C$ at the end.

Why This Matters for Exams

The antiderivative of a sum is the first tool for any integral that isn't a single standard form. It lets you break $\int (3x^2 + 2x + 1)\,dx$ into three easy integrals, or split $\int (\sin x + e^x)\,dx$ into known results.

Common mistake: trying to apply it to products or quotients. It does not work there: …

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Integrate each term: $\int(x+2)\,dx=\tfrac{x^2}{2}+2x+k$ .

Apply the power rule $\int x^n\,dx=\tfrac{x^{n+1}}{n+1}$ to each term:

…

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set A1 markMCQ
$\int(x + 2)\,dx =$
(a) $(x + 2)^3 + k$
(b) $\frac{x^2}{2} + k$
(c) $\frac{x^2}{2} + 2x + k$
(d) $\log(x + 2) + k$
›Reveal solutionSolution

Integrate each term: $\int(x+2)\,dx=\tfrac{x^2}{2}+2x+k$ .

Apply the power rule $\int x^n\,dx=\tfrac{x^{n+1}}{n+1}$ to each term:

…

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CBSE 2026Set ANNUAL1 mark
$\int(2x - 3\cos x + e^x)\,dx =$
›Reveal solutionSolution

Integrate term by term using standard integrals.

$\displaystyle\int(2x-3\cos x+e^x)\,dx = \int2x\,dx - 3\int\cos x\,dx+\int e^x\,dx$

…

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CBSE 2025Set E1 markMCQ
$\int_{0}^{1} (x + 2x + 3x^2 + 4x^3)\,dx =$
(a) $10$
(b) $\frac{5}{2}$
(c) $\frac{7}{2}$
(d) $\frac{1}{2}$
›Reveal solutionSolution

Combine $x+2x=3x$ ; integrate term by term over $[0,1]$ : $\frac32+1+1=\frac72$ .

First simplify: $x + 2x + 3x^2 + 4x^3 = 3x + 3x^2 + 4x^3.$ Integrate:

…

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CBSE 2024Set A11 markMCQ
$\int \sec x (\sec x + \tan x) \, dx$
(a) $\sec^2 x + \tan x + c$
(b) $\sec x + \tan x + c$
(c) $\sec x - \tan x + c$
(d) $-\tan x - \sec x + c$
›Reveal solutionSolution

Expand and integrate the standard forms to get $\tan x+\sec x+c$ , so (b). …

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CBSE 2024Set D1 markMCQ
$\int (x + \cos 2x)\,dx =$
(a) $\frac{1}{2}x\sin 2x + \frac{1}{4}\cos 2x + c$
(b) $\frac{1}{2}x\sin 2x - \frac{1}{4}\cos 2x + c$
(c) $2x\sin 2x + 4\cos 2x + c$
(d) $\frac{x^2}{2} + \frac{\sin 2x}{2} + c$
›Reveal solutionSolution

$\int(x+\cos 2x)\,dx=\frac{x^2}{2}+\frac{\sin 2x}{2}+c$ .

Integrate each term:

$\int x\,dx = \dfrac{x^2}{2}$ , and $\int \cos 2x\,dx = \dfrac{\sin 2x}{2}$ .

…

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CBSE 2024Set ANNUAL1 mark
$\displaystyle\int (2x - 3\cos x + e^x)\,dx$
›Reveal solutionSolution

Integrate each term separately using the standard formulas $\int x\,dx=\dfrac{x^2}{2}$ scaled by 2, $\int\cos x\,dx=\sin x$ , and $\int e^x\,dx=e^x$ .

$\displaystyle\int (2x - 3\cos x + e^x)\,dx = 2\int x\,dx - 3\int \cos x\,dx + \int e^x\,dx$

…

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CBSE 2023Set E1 markMCQ
$\displaystyle\int(4\cos x-5\sin x)\,dx =$
(a) $k+4\sin x+5\cos x$
(b) $k-4\sin x-5\cos x$
(c) $k+4\sin x-5\cos x$
(d) $k-4\sin x+5\cos x$
›Reveal solutionSolution

$\int4\cos x\,dx=4\sin x$ and $\int(-5\sin x)\,dx=+5\cos x$ , summing to $4\sin x+5\cos x+k$ .

Integrate each term:

$\int 4\cos x\,dx=4\sin x$ .

$\int (-5\sin x)\,dx=-5(-\cos x)=5\cos x$ .

…

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CBSE 2022Set M1 mark
$\int \sec x (\sec x + \tan x)\,dx$
›Reveal solutionSolution

Expand the integrand into standard forms $\sec^2x$ and $\sec x\tan x$ . …

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CBSE 2021Set I1 markMCQ
$\int (x+2)\,dx =$
(a) $(x+2)^3 + k$
(b) $\dfrac{x^2}{2} + k$
(c) $\dfrac{x^2}{2} + 2x + k$
(d) $\log(x+2) + k$
›Reveal solutionSolution

Integrate $x$ and $2$ separately.

…

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CBSE 2021Set I1 markMCQ
$\int_0^2 (x^2+1)\,dx =$
(a) $\dfrac{8}{3}$
(b) $\dfrac{14}{3}$
(c) $\dfrac{13}{3}$
(d) $\dfrac{1}{3}$
›Reveal solutionSolution

$\displaystyle\int_0^2 (x^2+1)\,dx = \dfrac{14}{3}$ .

Integrate term by term:

$\int_0^2 (x^2+1)\,dx = \left[\dfrac{x^3}{3} + x\right]_0^2.$

Evaluate at the limits: …

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CBSE 2020Set ANNUAL1 mark
$Q. Find \int(2x^2 + e^x) dx.$
›Reveal solutionSolution

$\displaystyle\int(2x^2+e^x)\,dx=\dfrac{2x^3}{3}+e^x+C$ .

Concept. Integration is linear, so we may integrate each term separately and pull out constants.

Step-by-step.

$\int 2x^2\,dx=2\cdot\frac{x^{3}}{3}=\frac{2x^3}{3},\qquad \int e^x\,dx=e^x.$ …

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