Question

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

• A. $\frac{1}{ln4}$
• B. $\frac{1}{{\left(ln2\right)}^2}$
• C. $\frac{1}{ln2}$
• D. $\frac{1}{{\left(ln4\right)}^2}$

Answer 1

Answer: (C)

Given, $I=\int 2^{2^x}.2^{x}dx$
Let, $2^{2^x}=t\Rightarrow 2^{2^x}{\left(.2\right)}^x.{\left(ln2\right)}^{2}dx=dt$
$\Rightarrow 2^{2^x}{\left(.2\right)}^{x}dx=\frac{dt}{{\left(ln2\right)}^2}$
Thus, $I=\int \frac{dt}{{\left(ln2\right)}^2}=\frac{t}{{\left(ln2\right)}^2}+C$
$=\frac{2^{2^x}}{{\left(ln2\right)}^2}+C$
$\therefore \lambda =\frac{1}{{\left(ln2\right)}^2}\Rightarrow \sqrt{\lambda }=\frac{l}{ln2}$