The integral $\int\left(\left(\frac{x}{2}\right)^{x}+\left(\frac{2}{x}\right)^{x}\right)\left(1+\log_{e}\frac{x}{2}\right) d x$ is equal to

(B) $\left(\frac{x}{2}\right)^{x}-\left(\frac{2}{x}\right)^{x}+C$

(A) $\left(\frac{x}{2}\right)^{x}+\left(\frac{2}{x}\right)^{x}+C$

(C) $\left(\frac{x}{2}\right)^{x} \log _{2}\left(\frac{x}{2}\right)+C$

(D) $\left(\frac{x}{2}\right)^{x} \log _{2}\left(\frac{2}{x}\right)+C$

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Indefinite Integrals

The integral

\int ((\frac{x}{2})^{x} + (\frac{2}{x})^{x}) (1 + lo g_{e} \frac{x}{2}) d x

is equal to

Answer: (B)

Let

u = (\frac{x}{2})^{x}

and

w = (\frac{2}{x})^{x}

. Then

\frac{u ^{'}}{u} = \frac{d}{d x} [x ln \frac{x}{2}] = 1 + ln \frac{x}{2}

and

\frac{w ^{'}}{w} = \frac{d}{d x} [x ln \frac{2}{x}] = - (1 + ln \frac{x}{2})

. Hence

\frac{d}{d x} [(\frac{x}{2})^{x} - (\frac{2}{x})^{x}] = (1 + ln \frac{x}{2}) [(\frac{x}{2})^{x} + (\frac{2}{x})^{x}]

, which is precisely the integrand. Therefore the integral equals

(\frac{x}{2})^{x} - (\frac{2}{x})^{x} + C

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Single correct option