The correct option is A a 2alog 2 To find : ∫_0^ax ax+a #10;· #160; dx #10;Consider, ∫x ax+a· #160; dx=∫ 1 2a ∫1(x+a)dx #10; ...

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Fundamental Laws of Logarithms

∫0ax-a/x+a dx...

Question

$\int_{0}^{a} \frac{x - a}{x + a} d x =$

a + 2 a log 2

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a - 2 a log 2

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2 a log - a

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2 a log 2

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Solution

The correct option is A $a - 2 a log 2$
To find : $\int_{0}^{a} \frac{x - a}{x + a} \cdot d x$
Consider, $\int \frac{x - a}{x + a} \cdot d x = \int 1 - 2 a \int \frac{1}{(x + a)} d x$
$= x - 2 a log (x + a) + c$
Then, $\int_{0}^{a} \frac{x - a}{x + a} \cdot d x = {[x - 2 a log (x + a) + c]}_{0}^{a}$
$= (a - 2 a log (1 a) + c) - [0 - 2 a log (a) + c]$
$= a - 2 a log 2$
$\int_{0}^{a} \frac{x - a}{x + a} \cdot d x = a - 2 a log 2$

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