Primitive of x squared over x squared minus a squared squared
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Theorem
- $\ds \int \frac {x^2 \rd x} {\paren {x^2 - a^2}^2} = \frac {-x} {2 \paren {x^2 - a^2} } + \frac 1 {4 a} \map \ln {\frac {x - a} {x + a} } + C$
for $x^2 > a^2$.
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 | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 {2 a} \paren {\frac {x - a} {x + a} } + a^2 \int \frac {\d x} {\paren {x^2 - a^2}^2} + C\) | Primitive of $\dfrac 1 {x^2 - a^2}$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 {2 a} \paren {\frac {x - a} {x + a} } + a^2 \paren {\frac {-x} {2 a^2 \paren {x^2 - a^2} } - \frac 1 {4 a^3} \paren {\frac {x - a} {x + a} } } + C\) | Primitive of $\dfrac 1 {\paren {x^2 - a^2}^2}$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {-x} {2 \paren {x^2 - a^2} } + \frac 1 {4 a} \map \ln {\frac {x - a} {x + a} } + C\) | simplifying |
$\blacksquare$
Also see
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $x^2 - a^2$, $x^2 > a^2$: $14.153$