The integral ∫ 1 2 e x . x x 2 + log e x dx equals :

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Question: Answered & Verified by Expert

The integral

\int_{1}^{2} e^{x} . x^{x} (2 + \log_{e} x) dx

equals :

MathematicsDefinite IntegrationJEE MainJEE Main 2020 (06 Sep Shift 2)

Options:

A $e (4 e + 1)$
B $4 e^{2} - 1$
C $e (4 e - 1)$
D $e (2 e - 1)$

Solution:

1441 Upvotes Verified Answer

The correct answer is:

e (4 e - 1)

Let $y = {(ex)}^{x}$

$l ny = x [1 + l nx]$

$\frac{1}{y} \frac{dy}{dx} = (2 + l nx)$

$\Rightarrow$ $dy = {(ex)}^{x} (2 + l nx) dx$

$\int_{1}^{2} e^{x} . x^{x} (2 + \log_{e} x) dx$

$= {(y)}_{1}^{2}$

$= {({(ex)}^{x})}_{1}^{2}$

$= 4 e^{2} - e$

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