Question

What is $ \int {{{\tan }^{ - 1}}} xdx? $

Answer 1

Hint: As we know that we have to integrate the above question. Also the question contains an inverse trigonometric function as tangent is a trigonometric ratio. In this question we will use the integration by parts formula or we can say partial integration. The formula of integration by parts is $ \int {udv = uv - \int {vdu} } $.

Complete step by step solution:
According to the question we have $ I = \int {{{\tan }^{ - 1}}(x)dx} $ .
Let us assume as $ u = {\tan ^{ - 1}}(x) $ , so we can say that $ du = 34{1}{{{x^2} + 1}}dx $ . Similarly we assume that $ dv = dx $ or we can say $ v = x $ .
Now the formula says that $ \int {udv = uv - \int {vdu} } $ .
By applying the above formula we can and putting the values we can write
$ I = {\tan ^{ - 1}}(x) \cdot x - \int {x \cdot \dfrac{1}{{{x^2} + 1}}dx} $ .
It can be further written as
$ I = x{\tan ^{ - 1}}(x) - \int {\dfrac{x}{{{x^2} + 1}}dx} $ .
We can make a substitution that $ u = {x^2} + 1 $ , it can be further written as
$ 2xdx = du \Rightarrow xdx = \dfrac{{du}}{2} $ .
By applying the above we have
$ I = x{\tan ^{ - 1}}x - \int {u \cdot xdx} $ , We can put the value of $ xdx $ ,
It further gives us
$ x{\tan ^{ - 1}}x - \dfrac{1}{2}\int {\dfrac{1}{u}du}\\
  \Rightarrow x{\tan ^{ - 1}}x - \dfrac{1}{2}\ln (u) + C $ .
By substituting back the value of $ u = {x^2} + 1 $ , We have c
Hence the required value of the above question is
 $ I = x{\tan ^{ - 1}}x - \dfrac{1}{2}\ln ({x^2} + 1) + C $ .
So, the correct answer is “ $ I = x{\tan ^{ - 1}}x - \dfrac{1}{2}\ln ({x^2} + 1) + C $ ”.

Note: Before solving this kind of question we should have the proper knowledge of integration by parts and their formulas. WE should always remember the important formulas of differentiation and integration. It should be noted that differentiation of $ {\tan ^{ - 1}}x $ is $ \dfrac{1}{{{x^2} + 1}} $