Primitive of x squared over x squared plus a squared squared

Theorem

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

Proof 1

Let:

 $\ds x$ $=$ $\ds a \tan \theta$ $\ds \leadsto \ \$ $\ds \frac {\d x} {\d \theta}$ $=$ $\ds a \sec^2 \theta$ Derivative of Tangent Function $\ds \leadsto \ \$ $\ds \int \frac {x^2 \rd x} {\paren {x^2 + a^2}^2}$ $=$ $\ds \int \frac {a^2 \tan^2 \theta a \sec^2 \theta} {\paren {a^2 \tan^2 \theta + a^2}^2} \rd \theta$ Integration by Substitution $\ds$ $=$ $\ds \int \frac {a^3 \tan^2 \theta \sec^2 \theta} {a^4 \sec^4 \theta} \rd \theta$ Difference of Squares of Secant and Tangent $\ds$ $=$ $\ds \frac 1 a \int \frac {\tan^2 \theta} {\sec^2 \theta} \rd \theta$ simplification $\ds$ $=$ $\ds \frac 1 a \int \tan^2 \theta \cos^2 \theta \rd \theta$ Definition of Real Secant Function $\ds$ $=$ $\ds \frac 1 a \int \frac {\sin^2 \theta} {\cos^2 \theta} \cos^2 \theta \rd \theta$ Definition of Real Tangent Function $\ds$ $=$ $\ds \frac 1 a \int \sin^2 \theta \rd \theta$ simplification $\ds$ $=$ $\ds \frac \theta {2 a} - \frac {\sin 2 \theta} {4 a} + C$ Primitive of Square of Sine Function $\ds$ $=$ $\ds \frac \theta {2 a} - \frac 1 {2 a} \frac {\tan \theta} {1 + \tan^2 \theta} + C$ Tangent Half-Angle Substitution for Sine $\ds$ $=$ $\ds -\frac x {2 \paren {x^2 + a^2} } + \frac 1 {2 a} \arctan \frac x a + C$ substituting for $\theta$

$\blacksquare$

Proof 2

 $\ds \int \frac {x^2 \rd x} {\paren {x^2 + a^2}^2}$ $=$ $\ds \int \frac {x^2 + a^2 - a^2} {\paren {x^2 + a^2}^2} \rd x$ $\ds$ $=$ $\ds \int \frac {x^2 + a^2} {\paren {x^2 + a^2}^2} \rd x - a^2 \int \frac {\d x} {\paren {x^2 + a^2}^2}$ Linear Combination of Primitives $\ds$ $=$ $\ds \int \frac {\d x} {x^2 + a^2} + a^2 \int \frac {\d x} {\paren {x^2 + a^2}^2}$ simplification $\ds$ $=$ $\ds \frac 1 a \arctan {\frac x a} + a^2 \int \frac {\d x} {\paren {x^2 + a^2}^2} + C$ Primitive of Reciprocal of $x^2 + a^2$ $\ds$ $=$ $\ds \frac 1 a \arctan {\frac x a} + a^2 \paren {\frac x {2 a^2 \paren {x^2 + a^2} } + \frac 1 {2 a^3} \arctan \frac x a} + C$ Primitive of Reciprocal of $\paren {x^2 + a^2}^2$ $\ds$ $=$ $\ds \frac {-x} {2 \paren {x^2 + a^2} } + \frac 1 {2 a} \arctan \frac x a + C$ simplifying

$\blacksquare$