MHTCET 11 Apr 2026 Shift 2 Paper

Concept:Simplify the integrand, then substitute $t = 2^{x}$ to reduce it to a standard square-root integral.Explanation:Simplify the expression inside the square root.Since $4^{x} = (2^{x})^{2}$, we have:$$\sqrt{4^{x}(4^{x}+4)} = \sqrt{(2^{x})^{2}(4^{x}+4)} = 2^{x}\sqrt{4^{x}+4}$$So the integral becomes:$$\int 2^{x}\sqrt{4^{x}+4}\,dx$$Let $t = 2^{x}$.Then $\frac{dt}{dx} = 2^{x}\log 2$, so $2^{x}dx = \frac{dt}{\log 2}$.Also, $4^{x} = t^{2}$.Hence the integral transforms to:$$\frac{1}{\log 2}\int \sqrt{t^{2}+4}\,dt$$Use the standard formula:$$\int \sqrt{t^{2}+a^{2}}\,dt = \frac{t}{2}\sqrt{t^{2}+a^{2}} + \frac{a^{2}}{2}\log\left|t+\sqrt{t^{2}+a^{2}}\right| + C$$Here $a^{2}=4$, so:$$\frac{1}{\log 2}\left[\frac{t}{2}\sqrt{t^{2}+4} + 2\log\left|t+\sqrt{t^{2}+4}\right|\right] + C$$Substitute $t = 2^{x}$ back:$$\frac{1}{\log 2}\left[\frac{2^{x}}{2}\sqrt{4^{x}+4} + 2\log\left|2^{x}+\sqrt{4^{x}+4}\right|\right] + C$$This matches the expression given in option A.Answer:Option A