Primitive of x by Arccotangent of x over a

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Theorem

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

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:

 $\displaystyle \frac {\d v} {\d x}$ $=$ $\displaystyle x$ $\displaystyle \leadsto \ \$ $\displaystyle v$ $=$ $\displaystyle \frac {x^2} 2$ Primitive of Power

Then:

 $\displaystyle \int x \arccot \frac x a \rd x$ $=$ $\displaystyle \frac {x^2} 2 \arccot \frac x a - \int \frac {x^2} 2 \paren {\frac {-a} {x^2 + a^2} } \rd x + C$ Integration by Parts $\displaystyle$ $=$ $\displaystyle \frac {x^2} 2 \arccot \frac x a + \frac a 2 \int \frac {x^2 \rd x} {x^2 + a^2} + C$ Primitive of Constant Multiple of Function $\displaystyle$ $=$ $\displaystyle \frac {x^2} 2 \arccot \frac x a + \frac a 2 \paren {x - a \arctan {\frac x a} } + C$ Primitive of $\dfrac {x^2} {x^2 + a^2}$ $\displaystyle$ $=$ $\displaystyle \frac {x^2} 2 \arccot \frac x a + \frac a 2 \paren {x - a \paren {\frac \pi 2 - \arccot {\frac x a} } } + C$ Sum of Arctangent and Arccotangent $\displaystyle$ $=$ $\displaystyle \frac {x^2 + a^2} 2 \arccot \frac x a + \frac {a x} 2 - \frac {a^2 \pi} 2 + C$ simplifying $\displaystyle$ $=$ $\displaystyle \frac {x^2 + a^2} 2 \arccot \frac x a + \frac {a x} 2 + C$ subsuming $\dfrac {a^2 \pi} 2$ into the arbitrary constant

$\blacksquare$