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Question Number 214915 by Spillover last updated on 23/Dec/24

	Answered by maths2 last updated on 23/Dec/24

	$x \to 1 - x; L e t A b e e t h e i n t e g r a l$ $= \int_{0}^{1} \frac{d x}{(1 + e^{- 2 + 4 x}) (5 + 2 x - 2 x^{2})} \Rightarrow$ $2 A = \int_{0}^{1} [\frac{1}{(5 + 2 x - 2 x^{2})} . (\frac{1}{1 + e^{- 2 + 4 x}} + \frac{1}{1 + e^{2 - 4 x}}) d x$ $= \int_{0}^{1} \frac{d x}{5 + 2 x - 2 x^{2}} . . e a s y f r o m H e a r$
	Answered by Spillover last updated on 24/Dec/24

	Commented by MathematicalUser2357 last updated on 25/Dec/24

	$What does KR mean ? Korea ?$
	Commented by TonyCWX08 last updated on 26/Dec/24

	${K i n g}^{'} s R u l e$ $\underset{a}{\int^{b}} f (x) d x = \underset{a}{\int^{b}} f (a + b - x) d x$
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