Question

Find the integral of the following function, the given function is \[\int {\dfrac{{dx}}{{{x^4} + 1}}} \]?

Answer 1

Hint: Integrating a function needs to solve the equation according to the basic formulae used in integration. The general formulae for integration for any variable and integrating with respect to that variable says that the power of the variable will increase by one and the final power on the variable will be multiplied in the denominator in the result obtained.
Formulae Used:
\[ \Rightarrow \int {\dfrac{{dx}}{{{x^2} + {a^2}}} = \dfrac{1}{a}{{\tan }^{ - 1}}\left( {\dfrac{x}{a}}
\right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \]
\[ \Rightarrow \]\[\int {\dfrac{{dx}}{{{x^2} + {a^2}}} = \dfrac{1}{{2a}}\ln |\dfrac{{x - a}}{{x + a}}|} \]

Complete step by step solution:
The given question is \[\int {\dfrac{{dx}}{{{x^4} + 1}}} \]
Here we have to multiply and divide by some variable and some constant numbers and then modify the equation to move further and then after using standard formulae we can get our final integral, on solving we get:
\[
\Rightarrow \int {\dfrac{{dx}}{{{x^4} + 1}}} \\
\Rightarrow \smallint \dfrac{1}{{1 + {x^4}}}{\text{ }}dx \\
\Rightarrow \int {\dfrac{{\dfrac{1}{{{x^2}}}}}{{\dfrac{{1 + {x^4}}}{{{x^2}}}}}dx} \\
\Rightarrow \dfrac{1}{2}\int {\dfrac{{\dfrac{2}{{{x^2}}}}}{{\dfrac{{1 + {x^4}}}{{{x^2}}}}}dx} \\
\Rightarrow \dfrac{1}{2}\int {\dfrac{{\left( {1 + \dfrac{1}{{{x^2}}}} \right) - \left( {1 -
\dfrac{1}{{{x^2}}}} \right)}}{{\dfrac{{1 + {x^4}}}{{{x^2}}}}}dx} \\
\Rightarrow \dfrac{1}{2}\int {\dfrac{{\left( {1 + \dfrac{1}{{{x^2}}}} \right)dx}}{{\dfrac{{1 +
{x^4}}}{{{x^2}}}}}} - \dfrac{1}{2}\int {\dfrac{{\left( {1 - \dfrac{1}{{{x^2}}}} \right)}}{{\dfrac{{1 +
{x^4}}}{{{x^2}}}}}} dx \\
\Rightarrow \dfrac{1}{2}\int {\dfrac{{\left( {1 + \dfrac{1}{{{x^2}}}} \right)dx}}{{{x^2} +
\dfrac{1}{{{x^2}}} - 2 + 2}}} - \dfrac{1}{2}\int
\dfrac{{\left( {1 - \dfrac{1}{{{x^2}}}} \right)dx}}{{{x^2} + \dfrac{1}{{{x^2}}} - 2 + 2}} \\
\\
\\
\]
\[
\Rightarrow \dfrac{1}{2}\int {\dfrac{{d\left( {1 + \dfrac{1}{{{x^2}}}} \right)}}{{{{\left( {x -
\dfrac{1}{x}} \right)}^2} + 2}}} - \dfrac{1}{2}\int {\dfrac{{d\left( {1 + \dfrac{1}{{{x^2}}}}
\right)}}{{{{\left( {x + \dfrac{1}{x}} \right)}^2} - 2}}} \\
\Rightarrow \dfrac{1}{2}\int {\dfrac{{d\left( {1 + \dfrac{1}{{{x^2}}}} \right)}}{{{{\left( {x -
\dfrac{1}{x}} \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}}}} - \dfrac{1}{2}\int {\dfrac{{d\left( {1 +
\dfrac{1}{{{x^2}}}} \right)}}{{{{\left( {x - \dfrac{1}{x}} \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}}}}
\\
u\sin g\,s\tan dard\,formulae\,:\,\int {\dfrac{{dx}}{{{x^2} + {a^2}}} = \dfrac{1}{a}{{\tan }^{ - 1}}\left(
{\dfrac{x}{a}} \right)\,and\,} \int {\dfrac{{dx}}{{{x^2} + {a^2}}} = \dfrac{1}{{2a}}\ln |\dfrac{{x - a}}{{x +
a}}|} \\
\Rightarrow \dfrac{1}{2} \times \dfrac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\dfrac{{x -
\dfrac{1}{x}}}{{\sqrt 2 }}} \right) - \dfrac{1}{2} \times \dfrac{1}{{2\sqrt 2 }}\ln |\dfrac{{x + \dfrac{1}{x}
- \sqrt 2 }}{{x + \dfrac{1}{x} + \sqrt 2 }}| + C \\
\Rightarrow \dfrac{1}{{2\sqrt 2 }}{\tan ^{ - 1}}\left( {\dfrac{{x - 1}}{{x\sqrt 2 }}} \right) -
\dfrac{1}{{4\sqrt 2 }}\ln |\dfrac{{x - x\sqrt 2 + 1}}{{x + x\sqrt 2 + 1}}| + C \\
\]
This is our final required integral we are seeking for.

Additional Information:
In the above question we have multiplied and divided certain variables and constants in the equation, and these variables and constants need to be divided for moving further, if you practice more and more then only you will get to how to adjust the equation as per our need.

Note: Finding integral of a function becomes little hard when the question need any specific formulae to go through, in this question also you have to do the steps same as they are done otherwise you will get to the result, integration is all about remembering the process for a particular kind of question, and you have to solve the question as per the question needs.