Class CBSE Class 12 Integrals Question

Question

If
$
\int \frac{2^\frac{1}{x}}{x^2}  dx = k \cdot 2^{\frac{1}{x}} + C,
$
then $k$ is equal to

Accepted Answer

Correct Answer: A. Explanation:

Let $u = \frac{1}{x} = x^{-1}$

Differentiating $u$ with respect to $x$ : $ du = -\frac{1}{x^2} dx$

Substitute $u$ and $du$ into the integral:

$\displaystyle \int \frac{1}{x^2} 2^{\frac{1}{x}} dx = \int 2^u (-du) = - \int 2^u du$

Using the standard integral formula $\displaystyle \int a^u du = \frac{a^u}{\ln a} + C$ (with $a=2$):

$\displaystyle - \int 2^u du = - \left( \frac{2^u}{\ln 2} \right) + C$

Hence

$\displaystyle \int \frac{1}{x^2} 2^{\frac{1}{x}} dx = - \frac{1}{\ln 2} \cdot 2^{\frac{1}{x}} + C$

Determining the Value of $k$

By comparing our result with the given form $\displaystyle k \cdot 2^{\frac{1}{x}} + C$:

$\displaystyle k \cdot 2^{\frac{1}{x}} + C = - \frac{1}{\ln 2} \cdot 2^{\frac{1}{x}} + C$

The value of $k$ is:

$\displaystyle k = - \frac{1}{\ln 2}$

Since $\log 2$ in the options typically denotes the natural logarithm $\ln 2$ in calculus, the answer is:

$\displaystyle \frac{-1}{\log 2}$