The integration formulas for 1/u2 a2 in Theorem 6.9.6are valid for u>a. Show that the

Chapter 6, Problem 67

(choose chapter or problem)

The integration formulas for $1 / \sqrt{u^{2}-a^{2}}$ in Theorem 6.9.6 are valid for $u>a$. Show that the following formula is valid for $u<-a$:

$\int \frac{d u}{\sqrt{u^{2}-a^{2}}}=-\cosh ^{-1}\left(-\frac{u}{a}\right)+c$

$\ln \left|u+\sqrt{u^{2}-a^{2}}\right|+c$

Equation Transcription:

$1/\sqrt{{u}^{2}\ -\ {a}^{2}}$

$u>a$

$u<-a$

$\int\limits_{\ }^{\ }\frac{du}{\sqrt{{u}^{2}\ -\ {a}^{2}}}=-{cosh}^{-1}\left({-\frac{u}{a}}\right)+c$

$ln|u+\sqrt{{u}^{2}-{a}^{2}}|+c$

Text Transcription:

1/sqrt u^2-a^2

u>a

u<-a

int frac d u / sqrt u^2-a^2=-cosh^-1 (- frac u / a)+c

Ln|u+sqrt u^2-a^2|+c

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