Question

Integral 2^1/x/x^2dx=k.2^1/x then K is equal to​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{\displaystyle\int\dfrac{2^{1/x}}{x^2}\,dx=k\cdot2^{1/x}+C}$

$\sf{\blue{Let\,\,\dfrac{1}{x}=t}}\\\sf{\implies\blue{-\dfrac{1}{x^2}\,dx=dt}}\\\sf{\implies\blue{\dfrac{1}{x^2}\,dx=-dt}}$

So,

$\tt{-\displaystyle\int2^{t}\,dt=k\cdot2^{1/x}+C}$

$\tt{\implies-\dfrac{2^{t}}{\ln(2)}+C=k\cdot2^{1/x}+C}$

$\tt{\implies-\dfrac{1}{\ln(2)}\cdot2^{1/x}+C=k\cdot2^{1/x}+C}$

$\tt{\implies-\dfrac{1}{\log_{e}(2)}\cdot2^{1/x}+C=k\cdot2^{1/x}+C}$

$\tt{\implies-\log_{2}(e)\cdot2^{1/x}+C=k\cdot2^{1/x}+C}$

On comparing,

$\sf{k=-log_{2}(e)}$