Primitive of Arccotangent of x over a over x squared

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Theorem

$\displaystyle \int \frac {\arccot \frac x a \rd x} {x^2} = \frac {-\arccot \frac x a} x + \frac 1 {2 a} \map \ln {\frac {x^2 + a^2} {x^2} } + C$


Proof

With a view to expressing the primitive in the form:

$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

\(\displaystyle u\) \(=\) \(\displaystyle \arccot \frac x a\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\d u} {\d x}\) \(=\) \(\displaystyle \frac {-a} {x^2 + a^2}\) Derivative of $\arccot \dfrac x a$


and let:

\(\displaystyle \frac {\d v} {\d x}\) \(=\) \(\displaystyle \frac 1 {x^2}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle v\) \(=\) \(\displaystyle \frac {-1} x\) Primitive of Power


Then:

\(\displaystyle \int \frac {\arccot \frac x a \rd x} {x^2}\) \(=\) \(\displaystyle \arccot \frac x a \paren {\frac {-1} x} - \int \paren {\frac {-1} x} \paren {\frac {-a} {x^2 + a^2} } \rd x + C\) Integration by Parts
\(\displaystyle \) \(=\) \(\displaystyle \frac {-\arccot \frac x a} x - a \int \frac {\d x} {x \paren {x^2 + a^2} } \rd x + C\) simplifying
\(\displaystyle \) \(=\) \(\displaystyle \frac {-\arccot \frac x a} x - \frac 1 a \paren {\frac 1 {2 a^2} \map \ln {\frac {x^2} {x^2 + a^2} } } + C\) Primitive of $\dfrac 1 {x \paren {x^2 + a^2} }$
\(\displaystyle \) \(=\) \(\displaystyle \frac {-\arccot \frac x a} x + \frac 1 {2 a} \map \ln {\frac {x^2 + a^2} {x^2} } + C\) Logarithm of Reciprocal

$\blacksquare$


Also see


Sources