Primitive of x squared over x squared plus a squared squared/Proof 2
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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
| \(\ds \int \frac {x^2 \rd x} {\paren {x^2 + a^2}^2}\) | \(=\) | \(\ds \int \frac {x^2 + a^2 - a^2} {\paren {x^2 + a^2}^2} \rd x\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \int \frac {x^2 + a^2} {\paren {x^2 + a^2}^2} \rd x - a^2 \int \frac {\d x} {\paren {x^2 + a^2}^2}\) | 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 |
$\blacksquare$