Primitive of x by Arctangent of x over a

From ProofWiki
Jump to navigation Jump to search

Theorem

$\ds \int x \arctan \frac x a \rd x = \frac {x^2 + a^2} 2 \arctan \frac x a - \frac {a x} 2 + C$


Proof

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:

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


and let:

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


Then:

\(\ds \int x \arctan \frac x a \rd x\) \(=\) \(\ds \frac {x^2} 2 \arctan \frac x a - \int \frac {x^2} 2 \paren {\frac a {x^2 + a^2} } \rd x + C\) Integration by Parts
\(\ds \) \(=\) \(\ds \frac {x^2} 2 \arctan \frac x a - \frac a 2 \int \frac {x^2 \rd x} {x^2 + a^2} + C\) Primitive of Constant Multiple of Function
\(\ds \) \(=\) \(\ds \frac {x^2} 2 \arctan \frac x a - \frac a 2 \paren {x - a \arctan \frac x a} + C\) Primitive of $\dfrac {x^2} {x^2 + a^2}$
\(\ds \) \(=\) \(\ds \frac {x^2 + a^2} 2 \arctan \frac x a - \frac {a x} 2 + C\) simplifying

$\blacksquare$


Also see


Sources