Question

How do you integrate \[\int {x{{\tan }^{ - 1}}x} \,\,dx\] by integration by parts method?

Answer 1

Hint: Here in this question given an indefinite integral, we have to find the integrated value of a given function. It can be solved by the method of integration by parts by separating the function as \[u\] and \[v\] , later integrated by using the standard formulas of integration. And by further simplification we get the required solution.

Complete step-by-step answer:
Integration by parts is a technique for performing indefinite integration or definite integration by expanding the differential of a product of functions \[d\left( {uv} \right)\] and expressing the original integral in terms of a known integral \[\int {v\,du} \] . A single integration by parts starts with
 \[d(uv) = u\,dv + v\,du\]
and integrates both sides,
 \[\int {d(uv)} = uv = \int {u\,dv} + \int {v\,} du.\] ------(1)
Rearranging gives
 \[\int {u\,} dv = uv - \int v \,du.\] ---------(2)
Consider the given function \[\int {x{{\tan }^{ - 1}}x} \,\,dx\] -----(3)
Given an integral which is not having any upper and lower limit then it’s an indefinite integral. Hence we add the C while integrating. Where, C is an arbitrary constant called as the constant of integration.
we can pick \[u\] as \[{\tan ^{ - 1}}x\] , because who knows the antiderivative of \[{\tan ^{ - 1}}x\] is \[\dfrac{1}{{1 + {x^2}}}\]
Then, of course \[dv = x\,dx\] .
i.e., \[u = {\tan ^{ - 1}}x\] \[ \Rightarrow \,\,\dfrac{{du}}{{dx}} = \dfrac{1}{{1 + {x^2}}}\] and
 \[dv = x\,dx\] \[ \Rightarrow \,\,v = \dfrac{{{x^2}}}{2}\]
Then by the method of integration by parts i.e., by the equation (2)
 \[ \Rightarrow \int {x{{\tan }^{ - 1}}x} \,\,dx = \,\,{\tan ^{ - 1}}x\dfrac{{{x^2}}}{2} - \int {\dfrac{{{x^2}}}{2}\dfrac{1}{{1 + {x^2}}}\,} dx\]
 \[ \Rightarrow \int {x{{\tan }^{ - 1}}x} \,\,dx = \,\,{\tan ^{ - 1}}x\dfrac{{{x^2}}}{2} - \dfrac{1}{2}\int {\dfrac{{{x^2}}}{{1 + {x^2}}}\,} dx\]
Add and subtract 1 in the numerator of second term in RHS
 \[ \Rightarrow \int {x{{\tan }^{ - 1}}x} \,\,dx = \,\,{\tan ^{ - 1}}x\dfrac{{{x^2}}}{2} - \dfrac{1}{2}\left[ {\int {\dfrac{{{x^2} + 1 - 1}}{{1 + {x^2}}}\,} } \right] dx\]
 \[ \Rightarrow \int {x{{\tan }^{ - 1}}x} \,\,dx = \,\,{\tan ^{ - 1}}x\dfrac{{{x^2}}}{2} - \dfrac{1}{2}\left[ {\int {\dfrac{{{x^2} + 1}}{{1 + {x^2}}}\,} dx - \int {\dfrac{1}{{1 + {x^2}}}\,} dx} \right] \]
On simplification, we get
 \[ \Rightarrow \int {x{{\tan }^{ - 1}}x} \,\,dx = \,\,{\tan ^{ - 1}}x\dfrac{{{x^2}}}{2} - \dfrac{1}{2}\left[ {\int {1\,} dx - \int {\dfrac{1}{{1 + {x^2}}}\,} dx} \right] \]
On integrating the second term of RHS, we get
 \[ \Rightarrow \int {x{{\tan }^{ - 1}}x} \,\,dx = \,\,{\tan ^{ - 1}}x\dfrac{{{x^2}}}{2} - \dfrac{1}{2}\left[ {x - {{\tan }^{ - 1}}x} \right] + C\]
\[ \Rightarrow \int {x{{\tan }^{ - 1}}x} \,\,dx = \,\,\dfrac{{{x^2}{{\tan }^{ - 1}}x}}{2} - \dfrac{x}{2} + \dfrac{{{{\tan }^{ - 1}}x}}{2} + C\]
Where C is an integrating constant.
Hence, the value of \[\int {x{{\tan }^{ - 1}}x} \,\,dx\] is \[\,\,\dfrac{{{x^2}{{\tan }^{ - 1}}x}}{2} - \dfrac{x}{2} + \dfrac{{{{\tan }^{ - 1}}x}}{2} + C\] .
So, the correct answer is “ \[\,\,\dfrac{{{x^2}{{\tan }^{ - 1}}x}}{2} - \dfrac{x}{2} + \dfrac{{{{\tan }^{ - 1}}x}}{2} + C\] ”.

Note: In integration we have two kinds one is definite integral and other one is indefinite integral. This question comes under the indefinite integral. While integrating the function which is in the form of product or division form we use the integration by parts method. By applying the integration by parts we obtain the solution.