Primitive of Reciprocal of x squared by x squared plus a squared
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Theorem
- $\ds \int \frac {\d x} {x^2 \paren {x^2 + a^2} } = -\frac 1 {a^2 x} - \frac 1 {a^3} \arctan \frac x a + C$
Proof
\(\ds \int \frac {\d x} {x^2 \paren {x^2 + a^2} }\) | \(=\) | \(\ds \int \paren {\frac 1 {a^2 x^2} - \frac 1 {a^2 \paren {x^2 + a^2} } } \rd x\) | Partial Fraction Expansion | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {a^2} \int \frac {\d x} {x^2} - \frac 1 {a^2} \int \frac {1 \rd x} {x^2 + a^2}\) | Linear Combination of Primitives | |||||||||||
\(\ds \) | \(=\) | \(\ds -\frac 1 {a^2 x} - \frac 1 {a^2} \int \frac {x \rd x} {x^2 + a^2} + C\) | Primitive of Power | |||||||||||
\(\ds \) | \(=\) | \(\ds -\frac 1 {a^2 x} - \frac 1 {a^2} \paren {\frac 1 a \arctan \frac x a} + C\) | Primitive of $\dfrac 1 {x^2 + a^2}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds -\frac 1 {a^2 x} - \frac 1 {a^3} \arctan \frac x a + C\) | simplifying |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $x^2 + a^2$: $14.130$
- 2009: Murray R. Spiegel, Seymour Lipschutz and John Liu: Mathematical Handbook of Formulas and Tables (3rd ed.) ... (previous) ... (next): $\S 17$: Tables of Special Indefinite Integrals: $(6)$ Integrals Involving $x^2 + a^2$: $17.6.6.$