The correct option is A 1/2(a^2 b^2)log|x^2 a^2/x^2 b^2|+c ∫ #160; x(x^a a^2)(x^2 b^2)dx= #160; 12∫ #160; dx^2(x^2 a^2)(x^2 b^2) ⇒ #160; ...

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Method of Difference

∫x/x2-a2x2-b2...

Question

$\int \frac{x}{(x^{2} - a^{2}) (x^{2} - b^{2})} d x =$

\frac{1}{2 (a^{2} - b^{2})} log | \frac{x^{2} - a^{2}}{x^{2} - b^{2}} | + c

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\frac{1}{2 (a^{2} - b^{2})} log | \frac{x^{2} - b^{2}}{x^{2} - a^{2}} | + c

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log | \frac{x^{2} - a^{2}}{x^{2} - b^{2}} | + c

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\frac{1}{(a^{2} - b^{2})} log | \frac{x^{2} - a^{2}}{x^{2} - b^{2}} | + c

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Solution

The correct option is A $\frac{1}{2 (a^{2} - b^{2})} log | \frac{x^{2} - a^{2}}{x^{2} - b^{2}} | + c$
$\int \frac{x}{(x^{a} - a^{2}) (x^{2} - b^{2})} d x = \frac{1}{2} \int \frac{d x^{2}}{(x^{2} - a^{2}) (x^{2} - b^{2})}$

$\Rightarrow \frac{1}{2} \int \frac{d t}{(t - a^{2}) (t - b^{2})}$

$\frac{1}{2 (a^{2} - b^{2}} \int [\frac{1}{(t - a^{2})} - \frac{1}{(t - b^{2})}] d t$
$= \frac{1}{2 (a^{2} - b^{2})} log (t - a^{2}) - log (t - b^{2}) + c$
$= \frac{1}{2 (a^{2} - b^{2})} log (\frac{x^{2} - a^{2}}{x^{2} - b^{2}}) + c$

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Similar questions

\frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2})} = k [\frac{a^{2}}{x^{2} + a^{2}} - \frac{b^{2}}{x^{2} + b^{2}}] \Rightarrow k =

x = \frac{\sqrt{a^{2} + b^{2}} + \sqrt{a^{2} - b^{2}}}{\sqrt{a^{2} + b^{2}} - \sqrt{a^{2} - b^{2}}}

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

\int \frac{x}{(x^{2} - a^{2}) (x^{2} - b^{2})} d x

\frac{1}{k (b^{2} - a^{2})} ln \frac{(x^{2} - b^{2})}{(x^{2} - a^{2})}

k

\in R^{+}

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - \frac{z^{2}}{c^{2}} = 1

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1

- \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - \frac{z^{2}}{c^{2}} = 1, \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1, - \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1

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