Class 12 Maths, Worked in Full Past papersGet updates
CBSE Board Examination 2025 · Section A · 1 mark

Q10: Find k if ∫21/xx2 dx=k⋅21/x+C\int \frac{2^{1/x}}{x^2}\,dx = k \cdot 2^{1/x} + C

Board 2025Ch 7 · Integrals
The question

If ∫21xx2 dx=k⋅21x+C\displaystyle\int \frac{2^{\frac{1}{x}}}{x^2}\,dx = k \cdot 2^{\frac{1}{x}} + C⁠, then k is equal to

(A) −1log⁡2\frac{-1}{\log 2}

(B) −log⁡2-\log 2

(C) −1-1

(D) 12\frac{1}{2}

Exam answer

Put 1x=t\frac{1}{x} = t⁠. Then −1x2 dx=dt-\frac{1}{x^2}\,dx = dt⁠, so dxx2=−dt\frac{dx}{x^2} = -dt⁠.

Why?

The exponent is 1x\frac{1}{x}⁠, and its derivative is −1x2-\frac{1}{x^2}⁠. Apart from the sign, that is the other factor of the integrand. So substituting t=1xt = \frac{1}{x} removes x completely:

ddx(x−1)=−x−2\frac{d}{dx}\left(x^{-1}\right) = -x^{-2}
∫21xx2 dx=−∫2t dt=−2tlog⁡2+C=−1log⁡2⋅21x+C\begin{aligned} \int \frac{2^{\frac{1}{x}}}{x^2}\,dx &= -\int 2^t\,dt \\ &= -\frac{2^t}{\log 2} + C \\ &= \frac{-1}{\log 2} \cdot 2^{\frac{1}{x}} + C \end{aligned}
Why?

Standard integral, for a>0a > 0⁠, a≠1a \ne 1⁠:

∫at dt=atlog⁡a+C\int a^t\,dt = \frac{a^t}{\log a} + C

It holds because the derivative of ata^t is atlog⁡aa^t \log a⁠. Then put back t=1xt = \frac{1}{x}⁠.

The first line, written out: split the integrand, then swap in t and −dt-dt⁠:

∫21xx2 dx=∫21x⋅dxx2=∫2t⋅(−dt)=−∫2t dt\begin{aligned} \int \frac{2^{\frac{1}{x}}}{x^2}\,dx &= \int 2^{\frac{1}{x}} \cdot \frac{dx}{x^2} \\ &= \int 2^t \cdot (-dt) \\ &= -\int 2^t\,dt \end{aligned}

Then use the standard integral with a=2a = 2⁠:

−∫2t dt=−2tlog⁡2+C-\int 2^t\,dt = -\frac{2^t}{\log 2} + C

Comparing with k⋅21x+Ck \cdot 2^{\frac{1}{x}} + C⁠, k=−1log⁡2k = \frac{-1}{\log 2}⁠.

Why?

Watch out: log⁡2\log 2 goes in the denominator. Multiplying by it instead gives −log⁡2-\log 2⁠, option (B). Dropping the minus sign from dtdt loses the sign of k.

Answer: (A) −1log⁡2\dfrac{-1}{\log 2}

Where this comes from

Chapter 7, Integrals: Integration by substitution.

Question type: Substitution

How to spot it: click to see

Type 2. Substitution: when the derivative is hiding in the question

Spot it: a function and its derivative both appear, like log⁡x\log x and 1x\frac{1}{x}⁠, or x2+1x^2 + 1 and xx⁠.

Here: The power 1x\frac1x has derivative −1x2-\frac{1}{x^2}⁠, and 1x2\frac{1}{x^2} is the other factor. Put 1x=t\frac1x = t⁠, so dxx2=−dt\frac{dx}{x^2} = -dt and only −∫2t dt-\int 2^t\,dt is left.

Quick lesson · Worked NCERT solutions · Interactive lesson

Found an error?

Tell us if something on this page is wrong or unclear. Send feedback on this question →

Integrals in past papers: question 7 of 46 · see them all