If $\int \frac{2^{1/x}}{x^2} dx = k \cdot 2^{1/x} + C$, then k is equal to (A) $\frac{-1}{\log 2}$ (B) $-\log 2$ (C) $-1$ (D) $\frac{1}{2}$

If ∫ frac21/xx² dx = k · 21/x + C, then k is equal to ( — CBSE Class 12 Mathematics 2025 PYQ

Question

\int \frac{2^{1/x}}{x^2} dx = k \cdot 2^{1/x} + C

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This problem asks for the coefficient $k$ in a given integral expression. We solve it by applying integration by parts to the left-hand side and comparing the resulting elementary term with the given form. The value of $k$ is $\boxed{-1}$ .

The problem asks us to find the value of $k$ given the equation $\int \frac{2^x}{x^2} dx = k \cdot 2^x \frac{1}{x} + C$ . The integral $\int \frac{2^x}{x^2} dx$ is a non-elementary integral, meaning it cannot be expressed in terms of elementary functions (polynomials, exponentials, logarithms, trigonometric functions).

In such problems, the given form $k \cdot 2^x \frac{1}{x} + C$ usually represents the elementary part obtained from a single application of integration by parts, with the remaining non-elementary integral implicitly absorbed or ignored for the purpose of finding $k$ . Our strategy will be to perform integration by parts on the left-hand side and then compare the resulting elementary term with $k \cdot 2^x \frac{1}{x}$ .

Recall the Integration by Parts Formula:

The integration by parts formula is given by:
$\int u \, dv = uv - \int v \, du$
Choose $u$ and $dv$ for the integral $\int \frac{2^x}{x^2} dx$ :

We have the integrand $\frac{2^x}{x^2} = 2^x \cdot x^{-2}$ .

To obtain a term like $2^x \frac{1}{x}$ in the $uv$ part, we should choose $dv$ such that $v$ involves $\frac{1}{x}$ .

Let's choose:
- $dv = x^{-2} dx$
- $u = 2^x$
Now, we find $v$ and $du$ :
- $v = \int x^{-2} dx = -x^{-1} = -\frac{1}{x}$
- $du = \frac{d}{dx}(2^x) dx = 2^x \log 2 \, dx$ (Recall that $\frac{d}{dx}(a^x) = a^x \log a$ )
Apply Integration by Parts:

Substitute these into the formula $\int u \, dv = uv - \int v \, du$ :

\int \frac{2^x}{x^2} dx = (2^x) \left(-\frac{1}{x}\right) - \int \left(-\frac{1}{x}\right) (2^x \log 2) dx

\int \frac{2^x}{x^2} dx = -\frac{2^x}{x} - \int \left(-\frac{2^x \log 2}{x}\right) dx

\int \frac{2^x}{x^2} dx = -\frac{2^x}{x} + \log 2 \int \frac{2^x}{x} dx

Compare with the given form: We are given that $\int \frac{2^x}{x^2} dx = k \cdot 2^x \frac{1}{x} + C$ . From our integration by parts, we found:

\int \frac{2^x}{x^2} dx = -\frac{2^x}{x} + \log 2 \int \frac{2^x}{x} dx

Comparing the elementary term involving $2^x \frac{1}{x}$: …

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

Showing the 12 most recent of 41 on this concept.

CBSE 2026Set A1 markMCQ
$\int_1^4\frac{dx}{\sqrt{x}} =$
(a) $1$
(b) $-2$
(c) $2$
(d) $-1$
›Reveal solutionSolution

$\int_1^4\tfrac{dx}{\sqrt{x}}=[2\sqrt{x}]_1^4=4-2=2$ .

Since $\int x^{-1/2}\,dx=2\sqrt{x}$ , evaluate between the limits:

…

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CBSE 2026Set ANNUAL1 mark
$3x^2+4x^3$
›Reveal solutionSolution

Apply the power rule for integration to each term separately.

$\int (3x^2+4x^3)\,dx = 3\int x^2\,dx + 4\int x^3\,dx$

$= 3\cdot\frac{x^3}{3} + 4\cdot\frac{x^4}{4} + C$

$= x^3 + x^4 + C$

…

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CBSE 2026Set ANNUAL1 mark
$\int_1^3 dx$
›Reveal solutionSolution

$\displaystyle\int_1^3 dx=[x]_1^3=3-1=2$ .

…

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CBSE 2025Set 65/1/11 markMCQ
$\int \frac{2^{1/x}}{x^2} dx = k \cdot 2^{1/x} + C$
›Reveal solutionSolution
This problem asks for the coefficient $k$ in a given integral expression. We solve it by applying integration by parts to the left-hand side and comparing the resulting elementary term with the given form. The value of $k$ is $\boxed{-1}$ .

The problem asks us to find the value of $k$ given the equation $\int \frac{2^x}{x^2} dx = k \cdot 2^x \frac{1}{x} + C$ . The integral $\int \frac{2^x}{x^2} dx$ is a non-elementary integral, meaning it cannot be expressed in terms of elementary functions (polynomials, exponentials, logarithms, trigonometric functions).

In such problems, the given form $k \cdot 2^x \frac{1}{x} + C$ usually represents the elementary part obtained from a single application of integration by parts, with the remaining non-elementary integral implicitly absorbed or ignored for the purpose of finding $k$ . Our strategy will be to perform integration by parts on the left-hand side and then compare the resulting elementary term with $k \cdot 2^x \frac{1}{x}$ .

Recall the Integration by Parts Formula:

The integration by parts formula is given by:
$\int u \, dv = uv - \int v \, du$

Choose $u$ and $dv$ for the integral $\int \frac{2^x}{x^2} dx$ :

We have the integrand $\frac{2^x}{x^2} = 2^x \cdot x^{-2}$ .

To obtain a term like $2^x \frac{1}{x}$ in the $uv$ part, we should choose $dv$ such that $v$ involves $\frac{1}{x}$ .

Let's choose:

$dv = x^{-2} dx$

$u = 2^x$

Now, we find $v$ and $du$ :

$v = \int x^{-2} dx = -x^{-1} = -\frac{1}{x}$

$du = \frac{d}{dx}(2^x) dx = 2^x \log 2 \, dx$ (Recall that $\frac{d}{dx}(a^x) = a^x \log a$ )

Apply Integration by Parts:

Substitute these into the formula $\int u \, dv = uv - \int v \, du$ :

$\int \frac{2^x}{x^2} dx = (2^x) \left(-\frac{1}{x}\right) - \int \left(-\frac{1}{x}\right) (2^x \log 2) dx$ $\int \frac{2^x}{x^2} dx = -\frac{2^x}{x} - \int \left(-\frac{2^x \log 2}{x}\right) dx$ $\int \frac{2^x}{x^2} dx = -\frac{2^x}{x} + \log 2 \int \frac{2^x}{x} dx$

Compare with the given form: We are given that $\int \frac{2^x}{x^2} dx = k \cdot 2^x \frac{1}{x} + C$ . From our integration by parts, we found:

$\int \frac{2^x}{x^2} dx = -\frac{2^x}{x} + \log 2 \int \frac{2^x}{x} dx$
Comparing the elementary term involving $2^x \frac{1}{x}$: …
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CBSE 2025Set X11 mark
$\int_{7}^{13} 1\,dx =$
›Reveal solutionSolution

$\displaystyle\int_{7}^{13} 1\,dx$ equals the interval length $13-7 = 6$ .

Concept: The definite integral of the constant function $1$ over $[a,b]$ equals $b-a$ .

Step 1 — Antiderivative.

$\int 1\,dx = x.$ …
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CBSE 2025Set ANNUAL1 mark
$\displaystyle\int \dfrac{1-x^3}{x^2}\,dx$
›Reveal solutionSolution

Split the integrand into two power-of- $x$ terms and integrate termwise.

…

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CBSE 2025Set E1 markMCQ
$\int_{0}^{1} 3x^2\,dx =$
(a) $3$
(b) $\frac{1}{3}$
(c) $1$
(d) $\frac{1}{9}$
›Reveal solutionSolution

Integrate the power and apply the limits; the value is $1$ .

…

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CBSE 2025Set E1 markMCQ
$\int x^m \cdot x^n\,dx =$
(a) $\frac{x^{m+1} \cdot x^{n+1}}{m + n + 2} + k$
(b) $\frac{x^{m+n}}{m + n} + k$
(c) $\frac{x^{m+n+1}}{m + n + 1} + k$
(d) $(m + n)x^{m+n-1} + k$
›Reveal solutionSolution

$x^m\cdot x^n = x^{m+n}$ , and $\int x^{m+n}\,dx = \frac{x^{m+n+1}}{m+n+1}+k$ .

Add the exponents:

$x^m\cdot x^n = x^{m+n}.$

Using the power rule $\int x^p\,dx = \frac{x^{p+1}}{p+1}+k$ with $p = m+n$ : …

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CBSE 2025Set E1 markMCQ
$\int \frac{(\sqrt{x} + 1)^2}{x\sqrt{x} + 2x + \sqrt{x}}\,dx =$
(a) $\sqrt{x} + k$
(b) $\frac{1}{2}\sqrt{x} + k$
(c) $2\sqrt{x} + k$
(d) $2x + k$
›Reveal solutionSolution

The integrand simplifies to $\frac{1}{\sqrt{x}}$ , whose integral is $2\sqrt{x}+k$ .

Factor the denominator:

$x\sqrt{x} + 2x + \sqrt{x} = \sqrt{x}\left(x + 2\sqrt{x} + 1\right) = \sqrt{x}\,(\sqrt{x} + 1)^2.$

So the integrand is …

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CBSE 2025Set E1 markMCQ
$\int_{0}^{a} \frac{dx}{\sqrt{x}} =$
(a) $2\sqrt{x}$
(b) $2\sqrt{a}$
(c) $\sqrt{x}$
(d) $\sqrt{a}$
›Reveal solutionSolution

$\int_0^a x^{-1/2}\,dx = [2\sqrt{x}]_0^a = 2\sqrt{a}$ .

Use the power rule $\int x^{-1/2}dx = 2x^{1/2}=2\sqrt{x}$ . Evaluating the definite integral:

…

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CBSE 2025Set E1 markMCQ
$2\int_{1}^{9} \frac{dx}{\sqrt{x}} =$
(a) $8$
(b) $4$
(c) $2$
(d) $12$
›Reveal solutionSolution

$2[2\sqrt{x}]_1^9 = 2(2\cdot3 - 2\cdot1) = 2\cdot4 = 8$ .

Use $\int x^{-1/2}dx = 2\sqrt{x}$ :

…

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CBSE 2025Set ANNUAL1 markMCQ
$\displaystyle\int x^{5/3}\, dx =$
(a) $\dfrac{3}{5}x^{2/3} + c$
(b) $\dfrac{8}{3}x^{8/3} + c$
(c) $\dfrac{3}{8}x^{8/3} + c$
(d) $\dfrac{5}{3}x^{8/3} + c$
›Reveal solutionSolution

Apply the standard power rule for integration, $\int x^n dx = \dfrac{x^{n+1}}{n+1}+c$ .

…

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Q.If $\int \frac{2^{1/x}}{x^2} dx = k \cdot 2^{1/x} + C$ , then k is equal to
(A) $\frac{-1}{\log 2}$
(B) $-\log 2$
(C) $-1$
(D) $\frac{1}{2}$

Concept understanding — Power Rule Integration

The Power Rule for Integration

The statement

Why it works

Using it

The one exception: $n=-1$

Q.If ∫21/xx2dx=k⋅21/x+C\int \frac{2^{1/x}}{x^2} dx = k \cdot 2^{1/x} + C∫x221/x​dx=k⋅21/x+C, then k is equal to (A) −1log⁡2\frac{-1}{\log 2}log2−1​ (B) −log⁡2-\log 2−log2 (C) −1-1−1 (D) 12\frac{1}{2}21​

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Q.If $\int \frac{2^{1/x}}{x^2} dx = k \cdot 2^{1/x} + C$ , then k is equal to
(A) $\frac{-1}{\log 2}$
(B) $-\log 2$
(C) $-1$
(D) $\frac{1}{2}$

Unlock everything free for 14 days