Question

How do you integrate $x\times {{\tan }^{-1}}x$ ?

Answer 1

Hint: Integrating $x\times {{\tan }^{-1}}x$ means$\int{x\times {{\tan }^{-1}}xdx}$ ,
We use the following four formulae:
Integration by parts: \[\underset{{}}{\overset{{}}{\mathop \int }}\,uvdx~=~u\underset{{}}{\overset{{}}{\mathop \int }}\,vdx- \underset{{}}{\overset{{}}{\mathop \int }}\,\left(\dfrac{du}{dx}\underset{{}}{\overset{{}}{\mathop \int }}\,vdx \right)dx....(i)\]
$\dfrac{d{{\tan }^{-1}}x}{dx}=\dfrac{1}{({{x}^{2}}+1)}......(ii)$
$\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+C....(iii)$ , where C is a constant
$\int{\dfrac{1}{{{x}^{2}}+1}dx={{\tan }^{-1}}dx+C}....(iv)$, where C is a constant
We first use (i) with u as ${{\tan }^{-1}}x$ and v as x, then use (ii) to find $\dfrac{du}{dx}$ and then use (iii) and (iv) to do further integration.

Complete step by step answer:
We first use integration by parts, which is \[\underset{{}}{\overset{{}}{\mathop \int }}\,uvdx~=~u\underset{{}}{\overset{{}}{\mathop \int }}\,vdx-\left( \underset{{}}{\overset{{}}{\mathop \int }}\,\dfrac{du}{dx}\underset{{}}{\overset{{}}{\mathop \int }}\,vdx \right)dx....(i)\]
for which we need the formula of differentiation of ${{\tan }^{-1}}x$ , which is $\dfrac{d{{\tan }^{-1}}x}{dx}=\dfrac{1}{({{x}^{2}}+1)}......(ii)$
So, applying (i) to $\int{x\times {{\tan }^{-1}}x}dx$ ,we have
$\Rightarrow \int{x\times {{\tan }^{-1}}xdx={{\tan }^{-1}}x\int{xdx-\int{\left( \dfrac{d{{\tan }^{-1}}x}{dx}\left( \int{xdx} \right)dx \right)}}}$
Using $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+C....(iii)$ , where C is a constant, we have:
$\Rightarrow {{\tan }^{-1}}x\int{xdx-\int{\left( \dfrac{d{{\tan }^{-1}}x}{dx}\left( \int{xdx} \right)dx \right)}}={{\tan }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)+{{C}_{1}}-\int{\dfrac{d{{\tan }^{-1}}x}{dx}\left( \dfrac{{{x}^{2}}}{2} \right)}dx$ , (Where ${{C}_{1}}$ is a constant)
Now using (ii)
$\Rightarrow {{\tan }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)+{{C}_{1}}-\int{\dfrac{d{{\tan }^{-1}}x}{dx}\left( \dfrac{{{x}^{2}}}{2} \right)}dx={{\tan }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)+{{C}_{1}}-\int{\dfrac{1}{{{x}^{2}}+1}\left( \dfrac{{{x}^{2}}}{2} \right)}dx$
Since constant can be taken out of the integration without changing its value, we have:
$\Rightarrow {{\tan }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)+{{C}_{1}}-\int{\dfrac{1}{{{x}^{2}}+1}\left( \dfrac{{{x}^{2}}}{2} \right)}dx={{\tan }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)+{{C}_{1}}-\dfrac{1}{2}\int{\dfrac{{{x}^{2}}}{{{x}^{2}}+1}}dx$
Now, since +1 -1 =0, we can add it to any term and fractions numerator. Thus,
$\Rightarrow {{\tan }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)+{{C}_{1}}-\dfrac{1}{2}\int{\dfrac{{{x}^{2}}}{{{x}^{2}}+1}}dx={{\tan }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)+{{C}_{1}}-\dfrac{1}{2}\int{\dfrac{{{x}^{2}}+1-1}{{{x}^{2}}+1}}dx$
We, know numerator distributes over a denominator, so \[\dfrac{{{x}^{2}}+1-1}{{{x}^{2}}+1}=\dfrac{{{x}^{2}}+1}{{{x}^{2}}+1}-\dfrac{1}{{{x}^{2}}+1}\]
Since ${{x}^{2}}$ is positive for all x, ${{x}^{2}}+1\ne 0$ . Therefore \[\dfrac{{{x}^{2}}+1}{{{x}^{2}}+1}-\dfrac{1}{{{x}^{2}}+1}=1-\dfrac{1}{{{x}^{2}}+1}\] as ${{x}^{2}}+1$ gets cancelled.
Thus, $\Rightarrow {{\tan }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)+{{C}_{1}}-\dfrac{1}{2}\int{\dfrac{{{x}^{2}}}{{{x}^{2}}+1}}dx={{\tan }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)+{{C}_{1}}-\dfrac{1}{2}\int{1-\dfrac{1}{{{x}^{2}}+1}}dx$
Integration distributes over addition that is, for any functions $f\text{ and }g$ we have $\left( \int{f+g} \right)=(\int{f)+\left( \int{g} \right)}$
So, $\Rightarrow {{\tan }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)+{{C}_{1}}-\dfrac{1}{2}\int{1-\dfrac{1}{{{x}^{2}}+1}}dx={{\tan }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)+{{C}_{1}}-\dfrac{1}{2}\left( \int{1dx-\int{\dfrac{1}{{{x}^{2}}+1}}}dx \right)$
Using (iii) again we have:
(Where ${{C}_{2}}$ is a constant)
$\Rightarrow {{\tan }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)+{{C}_{1}}-\dfrac{1}{2}\left( \int{1dx-\int{\dfrac{1}{{{x}^{2}}+1}}}dx \right)={{\tan }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)+{{C}_{1}}-\dfrac{1}{2}\left( x+{{C}_{2}}-\int{\dfrac{1}{{{x}^{2}}+1}dx} \right)$
Now, using $\int{\dfrac{1}{{{x}^{2}}+1}dx={{\tan }^{-1}}dx+C}....(iv)$ , where C is a constant, we have:
(where ${{C}_{3}}$ is a constant)
$\Rightarrow {{\tan }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)+{{C}_{1}}-\dfrac{1}{2}\left( x+{{C}_{2}}-\int{\dfrac{1}{{{x}^{2}}+1}dx} \right)={{\tan }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)+{{C}_{1}}-\dfrac{1}{2}\left( x+{{C}_{2}}-{{\tan }^{-1}}x+{{C}_{3}} \right)$
Since ${{C}_{1}},{{C}_{2}},{{C}_{3}}$ are all constants, they can be clubbed together as one constant and we write it as + C

Therefore, $\int{x\times {{\tan }^{-1}}xdx}={{\tan }^{-1}}x\left( \left( \dfrac{{{x}^{2}}}{2} \right)+\dfrac{1}{2} \right)-\dfrac{1}{2}x+C$

Note: Be careful with putting constants after doing integrals, it is advised to just add a C as constant after all the calculation instead of writing ${{C}_{1}},{{C}_{2}}$ etc, have written above for your understanding. That is a common place to make mistakes. Other than that, one must be extremely careful while applying the formulae as it's common to miss a term or two.
Alternatively,
You can substitute
 $\begin{align}
  & x=\tan \theta \\
 & dx={{\sec }^{2}}\theta d\theta \\
 & \int{x{{\tan }^{-1}}xdx=\int{\theta \tan \theta {{\sec }^{2}}\theta d\theta =\int{\dfrac{\theta \sin \theta }{{{\cos }^{3}}\theta }}}}d\theta \\
\end{align}$
And then use integration by parts.