Primitive of x by Arctangent of x

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Theorem

$\ds \int x \arctan x \rd x = \frac {x^2 + 1} 2 \arctan x - \frac x 2 + C$


Proof 1

From Primitive of $x \arctan \dfrac x a$:

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


The result follows on setting $a = 1$.

$\blacksquare$


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:

\(\ds u\) \(=\) \(\ds \arctan x\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d x}\) \(=\) \(\ds \frac 1 {x^2 + 1}\) Derivative of $\arctan x$


and let:

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


Then:

\(\ds \int x \arctan x \rd x\) \(=\) \(\ds \frac {x^2} 2 \arctan x - \int \frac {x^2} 2 \paren {\frac 1 {x^2 + 1} } \rd x + C\) Integration by Parts
\(\ds \) \(=\) \(\ds \frac {x^2} 2 \arctan x - \frac 1 2 \int \frac {x^2 \rd x} {x^2 + 1} + C\) Primitive of Constant Multiple of Function
\(\ds \) \(=\) \(\ds \frac {x^2} 2 \arctan x - \frac 1 2 \paren {x - 1 \arctan x} + C\) Primitive of $\dfrac {x^2} {x^2 + a^2}$ with $a = 1$
\(\ds \) \(=\) \(\ds \frac {x^2 + 1} 2 \arctan x - \frac x 2 + C\) simplifying

$\blacksquare$


Sources

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