Question

The integral $I=\displaystyle \int 2^{\left(2^{x} + x\right) }dx=\lambda .\left(2^{2^{x}}\right)+C$ (where, $C$ is the constant of integration). Then the value of $\sqrt{\lambda }$ is equal to

• A. $\frac{1}{ln 4}$
• B. $\frac{1}{\left(ln 2\right)^{2}}$
• C. $\frac{1}{ln 2}$
• D. $\frac{1}{\left(ln 4\right)^{2}}$

Answer 1

Answer: (C)

Given, $I=\displaystyle \int 2^{2^{x}}. 2^{x}dx$
Let, $2^{2^{x}}=t\Rightarrow 2^{2^{x}}\left(. 2\right)^{x}.\left(ln 2\right)^{2}dx=dt$
$\Rightarrow 2^{2^{x}}\left(. 2\right)^{x}dx=\frac{d t}{\left(ln 2\right)^{2}}$
Thus, $I=\displaystyle \int \frac{d t}{\left(ln 2\right)^{2}}=\frac{t}{\left(ln ⁡ 2\right)^{2}}+C$
$=\frac{2^{2^{x}}}{\left(ln 2\right)^{2}}+C$
$\therefore \lambda =\frac{1}{\left(ln 2\right)^{2}}\Rightarrow \sqrt{\lambda }=\frac{l}{ln ⁡ 2}$