Primitive of Arctangent Function/Proof 1
Theorem
- $\ds \int \arctan x \rd x = x \arctan x - \frac {\map \ln {x^2 + 1} } 2 + C$
Proof
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 |
$\blacksquare$