Primitive of Inverse Hyperbolic Tangent of x over a over x squared
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Theorem
- $\ds \int \frac 1 {x^2} \artanh \dfrac x a \rd x = -\frac 1 x \artanh \dfrac x a + \frac 1 {2 a} \map \ln {\frac {x^2} {a^2 - x^2} } + C$
Proof
With a view to expressing the primitive in the form:
- $\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
\(\ds u\) | \(=\) | \(\ds \artanh \frac x a\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d u} {\d x}\) | \(=\) | \(\ds \frac a {a^2 - x^2}\) | Derivative of $\artanh \dfrac x a$ |
and let:
\(\ds \frac {\d v} {\d x}\) | \(=\) | \(\ds \frac 1 {x^2}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds v\) | \(=\) | \(\ds \frac {-1} x\) | Primitive of Power |
Then:
\(\ds \int \frac 1 {x^2} \artanh \dfrac x a \rd x\) | \(=\) | \(\ds \paren {\artanh \frac x a} \paren {\frac {-1} x} - \int \paren {\frac {-1} x} \paren {\frac a {a^2 - x^2} } \rd x + C\) | Integration by Parts | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 x \artanh \dfrac x a + a \int \frac {\d x} {x \paren {a^2 - x^2} } + C\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 x \artanh \dfrac x a + a \paren {\frac 1 {2 a^2} \map \ln {\frac {x^2} {a^2 - x^2} } } + C\) | Primitive of $\dfrac 1 {x \paren {a^2 - x^2} }$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 x \artanh \dfrac x a + \frac 1 {2 a} \map \ln {\frac {x^2} {a^2 - x^2} } + C\) | simplifying |
$\blacksquare$
Also see
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving Inverse Hyperbolic Functions: $14.660$