The correct option is D log 2 ∫_0^111+xdx=ln(1+x)+c ∫_0^111+xdx = [ ln (1+x)+ c ]∫_0^1 =(ln(2)+c) (ln(1)+c) =ln2 ∫_0^111+xdx=ln2 ...

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Implicit Differentiation

∫011/1+xdx=

Question

$\int_{0}^{1} \frac{1}{1 + x} d x_{=}$

log 2

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\frac{1}{2} log 2

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2

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log 3

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Solution

The correct option is D $log 2$
$\int_{0}^{1} \frac{1}{1 + x} d x = ln (1 + x) + c$
$\int_{0}^{1} \frac{1}{1 + x} d x = [ln (1 + x) + c] \int_{0}^{1}$
$= (ln (2) + c) - (ln (1) + c)$
$= ln 2$
$\int_{0}^{1} \frac{1}{1 + x} d x = l n 2$

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