Question

Find the integral of ${\tan ^{ - 1}}\left( {2x} \right)$.

Answer 1

Hint:Indefinite integral simply represents the area under a given curve without any boundary conditions. So here by using this basic definition we can integrate ${\tan ^{ - 1}}\left( {2x} \right)dx$
Also we know integration by parts:
$\int {udv = uv - \int {vdu} } $
And: $\dfrac{d}{{dx}}{\tan ^{ - 1}}x = \dfrac{1}{{1 + {x^2}}}$
The above expressions can also be used to integrate ${\tan ^{ - 1}}\left( {2x} \right)$.

Complete step by step answer:
Given, ${\tan ^{ - 1}}\left( {2x} \right).....................\left( i \right)$.
Also by the basic definition of indefinite integral we can write that:
Indefinite integral is given by: $\int {f\left( x \right)dx} $
Such to integrate ${\tan ^{ - 1}}\left( {2x} \right)$ we can write
$\int {{{\tan }^{ - 1}}\left( {2x} \right)dx} ..........................\left( {ii} \right)$
Now on observing (i) we can say that the term ${\tan ^{ - 1}}\left( {2x} \right)$ cannot be integrated directly such that we have to use integration by parts, which is:
\[\int {udv = uv - \int {vdu} } ..................\left( {iii} \right)\]
Now here we need to find $u,du,v \;{\text{and}}\;dv$.
So from the given question we can write:
$u = {\tan ^{ - 1}}\left( {2x} \right)\;{\text{and}}\;dv = dx..................\left( {iv} \right)$
So we can find $v\;{\text{and}}\;du$from the given conditions in (iii):
\[u = {\tan ^{ - 1}}\left( {2x} \right)\;{\text{and}}\;dv = dx \\
\Rightarrow du = \dfrac{{d{{\tan }^{ - 1}}\left( {2x} \right)}}{{dx}}...............\left( v \right) \\ \]
Now to differentiate (v) we have the identity,
$\dfrac{d}{{dx}}{\tan ^{ - 1}}x = \dfrac{1}{{1 + {x^2}}}$
So applying the above identity to (v) we can write:
\[du = \dfrac{{d{{\tan }^{ - 1}}\left( {2x} \right)}}{{dx}} = \dfrac{2}{{1 + {{\left( {2x} \right)}^2}}}................\left( {vi} \right) \\
\Rightarrow v = x.......................\left( {vii} \right) \\ \]
Now substituting (iv), (vi) and (vii) in (iii), we get:
\[\int {udv = uv - \int {vdu} } \\
\Rightarrow\int {{{\tan }^{ - 1}}\left( {2x} \right)dx} = uv - \int {vdu} \\
\Rightarrow\int {{{\tan }^{ - 1}}\left( {2x} \right)dx = {{\tan }^{ - 1}}\left( {2x} \right) \times x - \int {x \times \left( {\dfrac{2}{{1 + {{\left( {2x} \right)}^2}}}} \right)dx} } \\
\Rightarrow\int {{{\tan }^{ - 1}}\left( {2x} \right)dx = x{{\tan }^{ - 1}}\left( {2x} \right) - 2\int {\left( {\dfrac{x}{{1 + 4{x^2}}}} \right)dx} } ..........................\left( {viii} \right) \\ \]
Now to integrate the given expression let us assume:
$u = 1 + 4{x^2} \\
\Rightarrow du = 8xdx \\
\Rightarrow xdx = \dfrac{1}{8}du \\ $
Now let’s consider the part \[\int {\left( {\dfrac{x}{{1 + 4{x^2}}}} \right)dx} \] and substitute the above assumptions:
Such that:
\[\int {\left( {\dfrac{x}{{1 + 4{x^2}}}} \right)dx} = \dfrac{1}{8}\int {\dfrac{1}{u}} du \\
\Rightarrow\int {\left( {\dfrac{x}{{1 + 4{x^2}}}} \right)dx} = \dfrac{1}{8}\ln \left( u \right) + C \\ \]
Now substitute the value of $u$ back in the equation we get:
\[\int {\left( {\dfrac{x}{{1 + 4{x^2}}}} \right)dx} = \dfrac{1}{8}\ln \left( {1 + 4{x^2}} \right) + C......................................\left( {ix} \right)\]
Now substituting it back to (viii), we can write:
\[\int {{{\tan }^{ - 1}}\left( {2x} \right)dx} = x{\tan ^{ - 1}}\left( {2x} \right) - \dfrac{2}{8}\ln \left( {1 + 4{x^2}} \right) + C \\
\therefore\int {{{\tan }^{ - 1}}\left( {2x} \right)dx} = x{\tan ^{ - 1}}\left( {2x} \right) - \dfrac{1}{4}\ln \left( {1 + 4{x^2}} \right) + C....................\left( x \right) \\ \]
Therefore our final answer is: \[x{\tan ^{ - 1}}\left( {2x} \right) - \dfrac{1}{4}\ln \left( {1 + 4{x^2}} \right) + C\]

Note:Since the basic definition indefinite integral simply implies the area under a curve such that the value of an integral must be finite or else the integral doesn’t exist.Similar to the above question if we can’t integrate an expression directly then we have to use the formula for integration by parts. Also we must know some basic identities of both integration as well as of derivatives.