Mathematics · 2021
JEE Main · 4 August 2021, Shift 2 · Q70
The value of the integral ∫_1^6 (log_e x d x)/(log_e x^2+ log_e(x^2-14 x+49)) is
The value of the integral $\displaystyle \int_1^6 \frac{\log _e x d x}{\log _e x^2+\log _e\left(x^2-14 x+49\right)}$ is
Official answer
(1)
$\displaystyle \frac{5}{4}$
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JEE Main 2021 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.