
How do you evaluate the integral $\int {\dfrac{{x - 1}}{{x + 1}}dx} $?
Answer
444.9k+ views
Hint: Start by splitting the integral so that it comes down to form which we can integrate easily. If there are any constants then get them out of the integral. Then mention all the standard integrands and use them to solve the integral.
Complete step by step solution:
First we will start off by splitting the integrand function.
\[
= \int {\dfrac{{x - 1}}{{x + 1}}dx} \\
= \int {\dfrac{{x + 1 - 2}}{{x + 1}}dx} \\
= \int {\left( {1 - \dfrac{2}{{x + 1}}} \right)dx} \\
\]
Now further we will divide the integral over the signs.
$\int {1dx - \int {\dfrac{2}{{x + 1}}dx} } $
Now we will use the standard integrals such as $\int {1dx} = x + c$ and $\int {\dfrac{1}{x}dx = \ln x}
$.
Now we substitute these integrals in our expression.
Also, the constant here which is $2$ will get out of the integral.
$
= \int {1dx - \int {\dfrac{2}{{x + 1}}dx} } \\
= \int {1dx - 2\int {\dfrac{1}{{x + 1}}dx} } \\
= x - 2(\ln |x + 1|) + c \\
$
Hence, the value of the integral $\int {\dfrac{{x - 1}}{{x + 1}}dx} $ will be $x - 2(\ln |x + 1|) + c$.
Additional Information: A derivative is the rate of change of a function with respect to a variable. Derivatives are fundamental to the solution of problems in calculus and differential equations. In general, scientists observe changing systems to obtain rate of change of some variable of interest, incorporate this information into some differential equation, and use integration techniques to obtain a function that can be used to predict the behaviour of the original system under diverse conditions.
Note: While substituting the terms make sure you are taking into account the signs of the terms as well. While using any standard integral check if your integral satisfies all the required conditions. While splitting the interval make sure you split terms along with their respective signs.
Complete step by step solution:
First we will start off by splitting the integrand function.
\[
= \int {\dfrac{{x - 1}}{{x + 1}}dx} \\
= \int {\dfrac{{x + 1 - 2}}{{x + 1}}dx} \\
= \int {\left( {1 - \dfrac{2}{{x + 1}}} \right)dx} \\
\]
Now further we will divide the integral over the signs.
$\int {1dx - \int {\dfrac{2}{{x + 1}}dx} } $
Now we will use the standard integrals such as $\int {1dx} = x + c$ and $\int {\dfrac{1}{x}dx = \ln x}
$.
Now we substitute these integrals in our expression.
Also, the constant here which is $2$ will get out of the integral.
$
= \int {1dx - \int {\dfrac{2}{{x + 1}}dx} } \\
= \int {1dx - 2\int {\dfrac{1}{{x + 1}}dx} } \\
= x - 2(\ln |x + 1|) + c \\
$
Hence, the value of the integral $\int {\dfrac{{x - 1}}{{x + 1}}dx} $ will be $x - 2(\ln |x + 1|) + c$.
Additional Information: A derivative is the rate of change of a function with respect to a variable. Derivatives are fundamental to the solution of problems in calculus and differential equations. In general, scientists observe changing systems to obtain rate of change of some variable of interest, incorporate this information into some differential equation, and use integration techniques to obtain a function that can be used to predict the behaviour of the original system under diverse conditions.
Note: While substituting the terms make sure you are taking into account the signs of the terms as well. While using any standard integral check if your integral satisfies all the required conditions. While splitting the interval make sure you split terms along with their respective signs.
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