Question

Evaluate the value of the given integral \[\int {\dfrac{{x - 1}}{{x + 1}}} dx\].

Answer 1

Hint: To obtain the value of this integral we need to reduce and simplify integrand , so that we can integrate easily . We can convert numerator with the help of denominator . Here we can write the numerator as a denominator minus some constant .
FORMULA USED:
\[\int {\dfrac{1}{x}} \,dx = \ln \,x\, + c\]
\[\int {dx\, = x + c} \]

Complete step-by-step answer:
First we will write the numerator in terms of the denominator .
\[x - 1\]= \[x + 1 - 2\]
Now we must rewrite the integral
\[\int {\dfrac{{x - 1}}{{x + 1}}} dx\]
\[ = \int {\dfrac{{x + 1 - 2}}{{x + 1}}} \,dx\]
After splitting the numerator we get
\[ = \int {\left( {1 - \dfrac{2}{{x + 1}}} \right)} \,dx\]
We know integration of addition or subtraction of two or more functions is equal to addition or subtraction of integration of those two or more functions .
\[ = \int {dx - \int {\dfrac{2}{{x + 1}}\,dx\,} } \]
\[ = \int {dx} - 2\int {\dfrac{1}{{x + 1}}} \,dx\]
After using basic integration rule we get
\[ = x - 2\,\ln \left| {x + 1} \right| + c\]
\[c\]=integration constant

Note: We have to keep in mind the basic integration rule and basic integration properties. We have to simplify the integrand as much as possible so that we can easily solve the integration. Otherwise integration will become complex and sometimes we will not be able to solve the integration at all . We can add integration constant for each part of the given integration when we integrate those parts . These constants will not affect the function in answer , as if we take the derivative of the answer we will get back the integrand (derivative of constant is zero ). Addition of all the constants will ultimately give some other constant . So we can add only one constant at the end .