Question

Evaluate the following definite integrals:
\[\int\limits_0^1 {\dfrac{{1 - x}}{{1 + x}}dx} \]

Answer 1

Hint: Split the terms in the numerator of the definite integral and divide them to make the integration easily. Then apply the definite integral values to the obtained integration solution. So, use this concept to reach the solution of the problem.

Complete step-by-step answer:
Let \[I = \int\limits_0^1 {\dfrac{{1 - x}}{{1 + x}}dx} \]
Adding and subtracting 1 to the numerator, we have
\[
 \Rightarrow I = \int\limits_0^1 {\dfrac{{2 - 1 - x}}{{1 + x}}dx} \\
  \Rightarrow I = \int\limits_0^1 {\dfrac{{2 - \left( {1 + x} \right)}}{{1 + x}}dx} \\
\]
Splitting the terms in numerator, we get
\[
 \Rightarrow I = \int\limits_0^1 {\left( {\dfrac{2}{{1 + x}} - \dfrac{{1 + x}}{{1 + x}}} \right)dx} \\
 \Rightarrow I = \int\limits_0^1 {\left( {\dfrac{2}{{1 + x}} - 1} \right)dx} \\
 \Rightarrow I = \int\limits_0^1 {\dfrac{2}{{1 + x}}dx} - \int\limits_0^1 {1dx} \\
\]
By integrating, we get
\[
 \Rightarrow I = \left[ {2\log \left( {1 + x} \right)} \right]_0^1 - \left[ x \right]_0^1 \\
 \Rightarrow I = 2\left[ {\log \left( {1 + x} \right)} \right]_0^1 - \left[ x \right]_0^1 \\
\]
Substituting the integral values, we get
\[
 \Rightarrow I = 2\left[ {\log \left( {1 + 1} \right) - \log \left( {1 + 0} \right)} \right] - \left[ {1 - 0} \right] \\
 \Rightarrow I = 2\left[ {\log \left( 2 \right) - \log \left( 1 \right)} \right] - \left[ {1 - 0} \right] \\
 \Rightarrow I = 2\left[ {\log 2 - 0} \right] - 1 \\
 \Rightarrow I = 2\log 2 - 1 \\
\]
Therefore, the solution of definite integral \[\int\limits_0^1 {\dfrac{{1 - x}}{{1 + x}}dx} \] is \[2\log 2 - 1\].

Note: A definite integral has start and end values: in other words there is an interval [a, b] where a and b are called limits, bounds or boundaries. Unlike in indefinite integration there will be no integrating constant in definite integration.