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## Homework Statement

I want to prove the standard result for indefinite integrals that [tex] \int \frac{1}{a^2-x^2} .dx\ =\ \frac{1}{2a}\ln{(\frac{a+x}{a-x})} \\ \mbox{But the problem is that i am unable to do so. I get this--}

\int {\frac{1}{a^2-x^2} .dx } \\ [/tex]

[tex]

= \frac{1}{2a}\int {\frac{2a}{a^2-x^2} .dx} \\

= \frac{1}{2a}\int{ \frac{a+x+a-x}{a^2-x^2}} \\ [/tex]

[tex]

= \frac{1}{2a}\int {\frac{1}{a-x}+\frac{1}{a+x} .dx} \\

[/tex]

[tex]

= \frac{1}{2a}(\int {\frac{1}{a-x}}+\int{\frac{1}{a+x}} ) \\

[/tex]

[tex]

= \frac{1}{2a}\ln{(a-x)}+ln{(a+x)} .dx [/tex]

I used this from an earlier derivation of [tex]

\int{\frac{1}{x^2-a^2} } \mbox{where } \frac{1}{2a}(\int{ \frac{1}{x-a}}+\int{\frac{1}{x+a}} ) [/tex]

was converted to [tex]

= \frac{1}{2a}\ln{(x-a)}-ln{(x-a)} .dx [/tex] similarly

Plz suggest where i went wrong and suggest a way to get the desired proof

Thx

EDIT: There's some trouble with latex. The code i entered was -: (space to avoid repetition)

[i tex] \int \frac{1}{a^2-x^2} .dx\ =\ \frac{1}{2a}\ln{(\frac{a+x}{a-x})} \\ \mbox{But the problem is that i am unable to do so. I get this--}

\int {\frac{1}{a^2-x^2} .dx } \\

= \frac{1}{2a}\int {\frac{2a}{a^2-x^2} .dx} \\

= \frac{1}{2a}\int{ \frac{a+x+a-x}{a^2-x^2}} \\

= \frac{1}{2a}\int {\frac{1}{a-x}+\frac{1}{a+x} .dx} \\

= \frac{1}{2a}(\int {\frac{1}{a-x}}+\int{\frac{1}{a+x}} ) \\

= \frac{1}{2a}\ln{a-x}+ln{a+x} .dx mbox{I used this from an earlier derivation of }

\int{\frac{1}{x^2-a^2} } \ mbox{where } \frac{1}{2a}(\int{ \frac{1}{x-a}}+\int{\frac{1}{x+a}} )

mbox{ was converted to ln similarly } [/i tex]

I am unable to get the complete msg?

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