Primitive of Power of x by Arccotangent of x over a

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Theorem

$\ds \int x^m \arccot \frac x a \rd x = \frac {x^{m + 1} } {m + 1} \arccot \frac x a + \frac a {m + 1} \int \frac {x^{m + 1} \rd x} {x^2 + a^2}$


Proof 1

Recall:

\(\text {(1)}: \quad\) \(\ds \int x^m \arccot x \rd x\) \(=\) \(\ds \frac {x^{m + 1} } {m + 1} \arccot x + \frac 1 {m + 1} \int \frac {x^{m + 1} \rd x} {x^2 + 1}\) Primitive of $x^m \arccot x$


Then:

\(\ds \int x^m \arccot \frac x a \rd x\) \(=\) \(\ds \int a^m \paren {\dfrac x a}^m \arccot \frac x a \rd x\) manipulating into appropriate form
\(\ds \) \(=\) \(\ds a^m \int \paren {\dfrac x a}^m \arccot \frac x a \rd x\) Primitive of Constant Multiple of Function
\(\ds \) \(=\) \(\ds a^m \paren {\dfrac 1 {1 / a} \paren {\frac 1 {m + 1} \paren {\dfrac x a}^{m + 1} \arccot \frac x a + \frac 1 {m + 1} \int \paren {\dfrac x a}^{m + 1} \frac {\d x} {\paren {\dfrac x a}^2 + 1} } }\) Primitive of Function of Constant Multiple, from $(1)$
\(\ds \) \(=\) \(\ds \frac {x^{m + 1} } {m + 1} \arccot \frac x a + \frac a {m + 1} \int \frac {x^{m + 1} \rd x} {x^2 + a^2}\) simplifying

$\blacksquare$


Proof 2

With a view to expressing the primitive in the form:

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

let:

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


and let:

\(\ds \frac {\d v} {\d x}\) \(=\) \(\ds x^m\)
\(\ds \leadsto \ \ \) \(\ds v\) \(=\) \(\ds \frac {x^{m + 1} } {m + 1}\) Primitive of Power


Then:

\(\ds \int x^m \arccot \frac x a \rd x\) \(=\) \(\ds \frac {x^{m + 1} } {m + 1} \arccot \frac x a - \int \frac {x^{m + 1} } {m + 1} \paren {\frac {-a} {x^2 + a^2} } \rd x\) Integration by Parts
\(\ds \) \(=\) \(\ds \frac {x^{m + 1} } {m + 1} \arccot \frac x a + \frac a {m + 1} \int \frac {x^{m + 1} \rd x} {x^2 + a^2}\) Primitive of Constant Multiple of Function

$\blacksquare$


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