Question

What is the integral of $\arctan \left( x \right)$ ?

Answer 1

Hint: To integrate $\arctan \left( x \right)$ , we will denote it as $\int{ta{{n}^{-1}}\left( x \right)}$ . We have to rewrite this integral in the form $\int{ta{{n}^{-1}}\left( x \right)}\times 1$ . We have to use integration by parts to integrate this. Integration by parts is given as $\int{udv}=uv-\int{vdu}$ . We have to take $u={{\tan }^{-1}}x$ and $dv=1$ . Find du and v from this and substitute in the formula for integration. Then we have integrated the second term and simplify if required.

Complete step by step solution:
We have to integrate $\arctan \left( x \right)$ . We can write this as
$\int{ta{{n}^{-1}}\left( x \right)}$
This is same as $\int{ta{{n}^{-1}}\left( x \right)}\times 1...\left( i \right)$ .
Let us use integration by parts. Integration by parts is given as
$\int{udv}=uv-\int{vdu}...\left( ii \right)$
Let us compare the above form to the equation (i). We can see that $u={{\tan }^{-1}}x$ and $dv=1$ . We have to find v and du so that we can substitute these in the above formula.
Let us consider $dv=1$ . Let us integrate this with respect to x.
$v=\int{dv}=\int{1}dx=x...\left( iii \right)$
Now, let us consider $u={{\tan }^{-1}}x$ . we have to differentiate u with respect to x.
 $\dfrac{du}{dx}=\dfrac{d}{dx}\left( {{\tan }^{-1}}x \right)=\dfrac{1}{1+{{x}^{2}}}$
Let us find du from the above equation.
$du=\dfrac{1}{1+{{x}^{2}}}dx...\left( iv \right)$
Let us substitute the value of u, dv, (iii) and (iv) in equation (i).
$\Rightarrow \int{{{\tan }^{-1}}x\times 1}=x{{\tan }^{-1}}x-\int{\dfrac{x}{1+{{x}^{2}}}dx}...\left( v \right)$
Let us consider $t=1+{{x}^{2}}$ . Now, we have differentiate t with respect to x.
$\dfrac{dt}{dx}=\dfrac{d}{dx}\left( 1+{{x}^{2}} \right)$
We know that derivative of a constant is 0 and $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{\left( n-1 \right)}}$ . Hence, we can write the above equation as
$\begin{align}
  & \Rightarrow \dfrac{dt}{dx}=\dfrac{d}{dx}\left( 1 \right)+\dfrac{d}{dx}\left( {{x}^{2}} \right) \\
 & \Rightarrow \dfrac{dt}{dx}=0+2x \\
 & \Rightarrow dt=2xdx \\
\end{align}$
Let us now, multiply and divide the numerator of second term in equation (v) by 2.
$\Rightarrow \int{{{\tan }^{-1}}x\times 1}=x{{\tan }^{-1}}x-\dfrac{1}{2}\int{\dfrac{2x}{1+{{x}^{2}}}dx}$
Let us substitute the values of t and dt in the above equation.
$\Rightarrow \int{{{\tan }^{-1}}x\times 1}=x{{\tan }^{-1}}x-\dfrac{1}{2}\int{\dfrac{dt}{t}}$
We know that $\int{\dfrac{dx}{x}}=\log \left| x \right|+C$ . Therefore, we can write the above equation as
$\Rightarrow \int{{{\tan }^{-1}}x\times 1}=x{{\tan }^{-1}}x-\dfrac{1}{2}\log \left| t \right|+C$
Let us substitute the value of t in the above equation.
$\Rightarrow \int{{{\tan }^{-1}}x\times 1}=x{{\tan }^{-1}}x-\dfrac{1}{2}\log \left| 1+{{x}^{2}} \right|+C$
We know that a square function will always be positive. Therefore $1+{{x}^{2}}$ will always be positive. Hence, we can write the above equation as
 \[\begin{align}
  & \Rightarrow \int{{{\tan }^{-1}}x\times 1}=x{{\tan }^{-1}}x-\dfrac{1}{2}\log \left( 1+{{x}^{2}} \right)+C \\
 & \Rightarrow \int{{{\tan }^{-1}}x}=x{{\tan }^{-1}}x-\dfrac{1}{2}\log \left( 1+{{x}^{2}} \right)+C \\
\end{align}\]
Hence, the integral of $\arctan \left( x \right)$ is \[x{{\tan }^{-1}}x-\dfrac{1}{2}\log \left( 1+{{x}^{2}} \right)+C\] .

Note: We took $\int{ta{{n}^{-1}}\left( x \right)}\times 1$ rather than $\int{1\times ta{{n}^{-1}}\left( x \right)}$ since integration by parts follows ILATE rule. I stands for inverse functions, L stands for logarithmic functions, A stands for algebraic functions, T stands for trigonometric functions and E stands for exponent functions. We can see that in $\int{ta{{n}^{-1}}\left( x \right)}\times 1$ , ${{\tan }^{-1}}x$ is an inverse function and 1 falls under algebraic function.