The integral $\int_{1}^{2} e^{x} \cdot x^{2}\left(2+\log _{e} x\right) d x$ equals:

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Indefinite Integrals

The integral

\int_{1}^{2} e^{x} \cdot x^{2} (2 + lo g_{e} x) d x

equals:

Answer: (C)

The integrand is a perfect derivative in disguise. Taking

y = (e x)^{x}

, so that

ln y = x (1 + ln x)

, differentiation gives

\frac{d y}{d x} = (e x)^{x} (2 + ln x) = e^{x} x^{x} (2 + ln x)

, which is exactly the function being integrated. The integral is therefore just

[(e x)^{x}]_{1}^{2} = (2 e)^{2} - (e)^{1} = 4 e^{2} - e = e (4 e - 1)

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Single correct option

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