Primitive of Arctangent Function

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Theorem

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


Corollary

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


Proof 1

Let:

\(\ds u\) \(=\) \(\ds \arctan x\)
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \tan u\) \(=\) \(\ds x\) Definition of Real Arctangent
\(\text {(2)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \sec u\) \(=\) \(\ds \sqrt {1 + x^2}\) Difference of Squares of Secant and Tangent


Then:

\(\ds \int \arctan x \rd x\) \(=\) \(\ds \int u \sec^2 u \rd u\) Primitive of Function of Arctangent
\(\ds \) \(=\) \(\ds u \tan u + \ln \size {\cos u} + C\) Primitive of $x \sec^2 a x$ with $a = 1$
\(\ds \) \(=\) \(\ds u \tan u - \ln \size {\sec u} + C\) Logarithm of Reciprocal and Secant is Reciprocal of Cosine
\(\ds \) \(=\) \(\ds u x - \ln \size {\sec u} + C\) Substitution for $\tan u$ from $\paren 1$
\(\ds \) \(=\) \(\ds u x - \ln \size {\sqrt {1 + x^2} } + C\) Substitution for $\sec u$ from $\paren 2$
\(\ds \) \(=\) \(\ds x \arctan x - \ln \size {\sqrt {1 + x^2} } + C\) Substitution for $u$
\(\ds \) \(=\) \(\ds x \arctan x - \frac {\ln \size {x^2 + 1} } 2 + C\) Logarithm of Power
\(\ds \) \(=\) \(\ds x \arctan x - \frac {\map \ln {x^2 + 1} } 2 + C\) $x^2 + 1$ always positive

$\blacksquare$


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:

\(\ds \frac {\d v} {\d x}\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds v\) \(=\) \(\ds x\) Primitive of Constant


Then:

\(\ds \int \arctan x \rd x\) \(=\) \(\ds x \arctan x - \int x \paren {\frac 1 {x^2 + 1} } \rd x + C\) Integration by Parts
\(\ds \) \(=\) \(\ds x \arctan x - \int \frac {x \rd x} {x^2 + 1} + C\) rearranging
\(\ds \) \(=\) \(\ds x \arctan x - \paren {\frac 1 2 \ln \paren {x^2 + 1} } + C\) Primitive of $\dfrac x {x^2 + a^2}$, setting $a := 1$
\(\ds \) \(=\) \(\ds x \arctan x - \frac {\ln \paren {x^2 + 1} } 2 + C\) simplifying

$\blacksquare$


Also presented as

This result can also be presented as:

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


Also see


Sources

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