First slide

Introduction to definite integration

Question

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

Easy

A

$2 \ln 2 - 1$
B

$\ln 2$
C

$\ln 2 + 1$
D

$2 \ln 2 + 1$

Solution

$\int_{0}^{1} \frac{1 - x}{1 + x} d x = \int_{0}^{1} (1 - \frac{2 x}{1 + x}) d x$
$= \int_{0}^{1} [1 - 2 (1 - \frac{1}{1 + x})] d x$

$= {[- x + 2 \log (1 + x)]}_{0}^{1} = - 1 + 2 \log 2$

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