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JEE MAIN 2020
06-09-2020 S2
Question
$$ \text { The integral } \int_1^2 e^x \cdot x^x\left(2+\log _e x\right) d x \text { equals : } $$
Select the correct option:
A
e(2e – 1)
B
e(4e – 1)
C
4e2 – 1
D
e(4e + 1)
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\begin{aligned} & I=\int_1^2 e^x x^x\left(2+\log _e x\right) d x \\ & I=\int_1^2 e^x x^x\left[1+\left(1+\log _e x\right)\right] d x \\ & =\int_1^2 e^x\left[x^x+x^x\left(1+\log _e x\right)\right] d x \\ & =\left[e^x x^x\right]_1^2,\left\{\int e^x\left(f(x)+f^{\prime}(x)\right) d x=e^x f(x)+c\right\} \\ & =e^2 \times 4-e \times 1 \\ & =4 e^2-e \\ & =e(4 e-1)\end{aligned}$
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