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

Last updated date: 13th Jul 2024
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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.