Question

How to integrate\[\int{x{{\cot }^{-1}}xdx}\]?

Answer 1

Hint: In the given question, we have been asked to integrate the function. In order to solve the question, we need to integrate the given function by using integration by parts. In integration by parts method.
Integration by parts is as follows:
\[\Rightarrow \int{f\left( x \right)\times g\left( x \right)dx=F\left( x \right)\times g\left( x \right)-\int{F\left( x \right)\times g'\left( x \right)dx}}\]
Where F(x) is the primitive (antiderivative) of f(x) and g’(x) is the derivative of g(x).

Formula used:
Integration by parts is as follows:
\[\Rightarrow \int{f\left( x \right)\times g\left( x \right)dx=F\left( x \right)\times g\left( x \right)-\int{F\left( x \right)\times g'\left( x \right)dx}}\]
Where F(x) is the primitive (antiderivative) of f(x) and g’(x) is the derivative of g(x).

Complete step-by-step solution:
We have the given solution,
\[\Rightarrow \int{x{{\cot }^{-1}}xdx}\]
Let,
\[\Rightarrow I=\int{x\times {{\cot }^{-1}}xdx}\]
Using integration by parts, we get
\[\Rightarrow I={{\cot }^{-1}}x\int{xdx}-\int{\left( \dfrac{d}{dx}\left( {{\cot }^{-1}}x \right)\int{xdx} \right)dx}\]
Therefore,
\[\Rightarrow I={{\cot }^{-1}}x\times \dfrac{{{x}^{2}}}{2}-\int{\dfrac{-1}{1+{{x}^{2}}}\times \dfrac{{{x}^{2}}}{2}dx}\]
\[\Rightarrow I={{\cot }^{-1}}x\times \dfrac{{{x}^{2}}}{2}+\dfrac{1}{2}\int{\dfrac{{{x}^{2}}}{1+{{x}^{2}}}dx}\]
\[\Rightarrow I={{\cot }^{-1}}x\times \dfrac{{{x}^{2}}}{2}+\dfrac{1}{2}\int{\dfrac{{{x}^{2}}}{1+{{x}^{2}}}dx}\]
Simplifying it further, we get
\[\Rightarrow I={{\cot }^{-1}}x\times \dfrac{{{x}^{2}}}{2}+\dfrac{1}{2}\int{\dfrac{\left( 1+{{x}^{2}} \right)-1}{1+{{x}^{2}}}dx}\]
Therefore,
\[\Rightarrow I={{\cot }^{-1}}x\times \dfrac{{{x}^{2}}}{2}+\dfrac{1}{2}\int{\left( 1-\dfrac{1}{1+{{x}^{2}}} \right)dx}\]
Integrating the last part, we get
\[\Rightarrow I={{\cot }^{-1}}x\times \dfrac{{{x}^{2}}}{2}+\dfrac{1}{2}\left( x-{{\tan }^{-1}}x \right)+C\]
Thus,
\[\Rightarrow I=\dfrac{{{x}^{2}}}{2}{{\cot }^{-1}}x+\dfrac{x}{2}-\dfrac{1}{2}{{\tan }^{-1}}x+C\]
Therefore,
\[\Rightarrow \int{x{{\cot }^{-1}}xdx}=\dfrac{{{x}^{2}}}{2}{{\cot }^{-1}}x+\dfrac{x}{2}-\dfrac{1}{2}{{\tan }^{-1}}x+C\], it is the required solution.

Note: In order to integrate the given question by using the integration by parts, we should always remember the formula for integrating in integration by parts method. As it is usually seen that some of the students while solving the question they make mistakes while choosing f(x) and g(x). So make sure that you choose f(x) and g(x) carefully and correctly. This will make integration easier to solve. There are also many other methods for integration. These are integration by substitution and integration by partial fractions. You should always remember all the methods for integration so that we can easily choose which method is suitable for solving the particular type of question. We should do all the calculations carefully and explicitly to avoid making errors.