How do you integrate $ \dfrac{1}{{{x^2} + 4}} $ ?
Answer
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Hint: As we know that integration is the process of finding functions whose derivative is given and named anti-differentiation or integration. The function is called the anti-derivative or integral or primitive of a given function $ f(x) $ and $ C $ is known as the constant of integration or the arbitrary constant. Here we have to integrate it by using the method of completing the square in the denominator and then integrating using $ u $ -substitution by the derivative of “arctan( $ x $ )
Complete step-by-step answer:
We should write the above given question in the form of the arctangent integral: $ \int {\dfrac{1}{{{u^2} + 1}}} du = \arctan (u) + C $ .
To get $ 1 $ in the denominator, we should start by factoring:
$ \int {\dfrac{1}{{{x^2} + 4}}} dx = \int {\dfrac{1}{{4(\dfrac{{{x^2}}}{4} + 1)}}} dx $ .
It can be written as,
$ \int {\dfrac{1}{{\dfrac{{{x^2}}}{4} + 1}}} dx $ .
Now we have to make $ {u^2} = \dfrac{{{x^2}}}{4} $ , so we assume that $ u = \dfrac{x}{2} $ , which implies that $ du = \dfrac{1}{2}dx $ . We have
$ \dfrac{1}{4}\int {\dfrac{1}{{\dfrac{{{x^2}}}{4} + 1}}} dx = \dfrac{1}{2}\int {\dfrac{{\dfrac{1}{2}}}{{{{(\dfrac{x}{2})}^2} + 1}}} dx $ . Therefore, $ \dfrac{1}{2}\int {\dfrac{1}{{{u^2} + 1}}} du $ .
Now by applying the arctangent integral it further gives us:
$ \dfrac{1}{2}\int {\dfrac{1}{{{u^2} + 1}}} du = \dfrac{1}{2}\arctan (u) + C \Rightarrow \dfrac{1}{2}\arctan (\dfrac{x}{2}) + C $ .
Hence the required answer is $ \dfrac{1}{2}\arctan (\dfrac{x}{2}) + C $ .
So, the correct answer is “$ \dfrac{1}{2}\arctan (\dfrac{x}{2}) + C $”.
Note: We know that integration by parts with $ u = \arctan (x) $ and $ dvdx = 1 $ gives $ v = x $ . Here we should divide in the above question whether the solution needs u or v, and try the u-substitution and then if cannot be simplified more then integrate it by parts. And the final answer must be written in the original variable of integration. It should always have $ C $ , known as the constant of integration or arbitrary constant. We should always add $ + C $ as the end of the solution. The function $ f(x) $ is called the integrand and $ f(x)dx $ is known as the element of integration.
Complete step-by-step answer:
We should write the above given question in the form of the arctangent integral: $ \int {\dfrac{1}{{{u^2} + 1}}} du = \arctan (u) + C $ .
To get $ 1 $ in the denominator, we should start by factoring:
$ \int {\dfrac{1}{{{x^2} + 4}}} dx = \int {\dfrac{1}{{4(\dfrac{{{x^2}}}{4} + 1)}}} dx $ .
It can be written as,
$ \int {\dfrac{1}{{\dfrac{{{x^2}}}{4} + 1}}} dx $ .
Now we have to make $ {u^2} = \dfrac{{{x^2}}}{4} $ , so we assume that $ u = \dfrac{x}{2} $ , which implies that $ du = \dfrac{1}{2}dx $ . We have
$ \dfrac{1}{4}\int {\dfrac{1}{{\dfrac{{{x^2}}}{4} + 1}}} dx = \dfrac{1}{2}\int {\dfrac{{\dfrac{1}{2}}}{{{{(\dfrac{x}{2})}^2} + 1}}} dx $ . Therefore, $ \dfrac{1}{2}\int {\dfrac{1}{{{u^2} + 1}}} du $ .
Now by applying the arctangent integral it further gives us:
$ \dfrac{1}{2}\int {\dfrac{1}{{{u^2} + 1}}} du = \dfrac{1}{2}\arctan (u) + C \Rightarrow \dfrac{1}{2}\arctan (\dfrac{x}{2}) + C $ .
Hence the required answer is $ \dfrac{1}{2}\arctan (\dfrac{x}{2}) + C $ .
So, the correct answer is “$ \dfrac{1}{2}\arctan (\dfrac{x}{2}) + C $”.
Note: We know that integration by parts with $ u = \arctan (x) $ and $ dvdx = 1 $ gives $ v = x $ . Here we should divide in the above question whether the solution needs u or v, and try the u-substitution and then if cannot be simplified more then integrate it by parts. And the final answer must be written in the original variable of integration. It should always have $ C $ , known as the constant of integration or arbitrary constant. We should always add $ + C $ as the end of the solution. The function $ f(x) $ is called the integrand and $ f(x)dx $ is known as the element of integration.
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