int frac{x^2-4}{x^4+9 x^2+16} cdot ,d x=tan ^ | vedclass

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(D) Let $I = \int \frac{x^2-4}{x^4+9 x^2+16} \,d x$.Divide the numerator and denominator by $x^2$:$I = \int \frac{1-\frac{4}{x^2}}{x^2+9+\frac{16}{x^2

Vedclass · Accepted Answer

$4$

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(D) Let $I = \int \frac{x^2-4}{x^4+9 x^2+16} \,d x$.Divide the numerator and denominator by $x^2$:$I = \int \frac{1-\frac{4}{x^2}}{x^2+9+\frac{16}{x^2}} \,d x$.Rewrite the denominator as a perfect square:$I = \int \frac{1-\frac{4}{x^2}}{(x^2+\frac{16}{x^2})+9} \,d x = \int \frac{1-\frac{4}{x^2}}{(x+\frac{4}{x})^2 - 2(x)(\frac{4}{x}) + 9} \,d x$.$I = \int \frac{1-\frac{4}{x^2}}{(x+\frac{4}{x})^2 + 1} \,d x$.Let $t = x+\frac{4}{x}$,then $dt = (1-\frac{4}{x^2}) \,d x$.Substituting these into the integral:$I = \int \frac{dt}{t^2+1} = 	an^{-1}(t) + c$.Thus,$I = 	an^{-1}(x+\frac{4}{x}) + c$.Comparing this with $	an^{-1}(f(x)) + c$,we get $f(x) = x+\frac{4}{x}$.Therefore,$f(2) = 2 + \frac{4}{2} = 2 + 2 = 4$.

$\int \frac{x^2-4}{x^4+9 x^2+16} \cdot \,d x=\tan ^{-1}(f(x))+c$ (where $c$ is a constant of integration),then the value of $f(2)$ is

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