Question

$\int \frac{x^2-4}{x^4+24x^2+16}dx$ equals

• A. $\frac{1}{4}ta{n}^{-1}\left(\frac{(x^2+4)}{4x}\right)+c$
• B. $-\frac{1}{4}co{t}^{-1}\left(\frac{(x^2+4)}{x}\right)+c$
• C. $-\frac{1}{4}co{t}^{-1}\left(\frac{4(x^2+4)}{x}\right)+c$
• D. $\frac{1}{4}co{t}^{-1}\left(\frac{4(x^2+4)}{x}\right)+c$

Answer 1

The correct option is A $\frac{1}{4}ta{n}^{-1}\left(\frac{(x^2+4)}{4x}\right)+c$

$\int \frac{x^2-4}{x^4+24x^2+16}dx=\int \frac{x^2-4}{(x^2-8\sqrt{2}+12)(x^2+8\sqrt{2}+12)}dx$
$=\int (\frac{2+\sqrt{2}}{2\sqrt{2}(x^2+8\sqrt{2}+12)}+\frac{2-\sqrt{2}}{2\sqrt{2}(-x^2+8\sqrt{2}-12)})dx$
$=(\frac{1}{2}+\frac{1}{\sqrt{2}})\int \frac{dx}{(8\sqrt{2}+12)(1+\frac{x^2}{8\sqrt{2}+12})}+(-\frac{1}{2}+\frac{1}{\sqrt{2}})\int \frac{dx}{(-x^2+8\sqrt{2}-12)}$
$=(\frac{1}{2(8\sqrt{2}-12)}+\frac{1}{\sqrt{2}(8\sqrt{2}+12)})\int \frac{dx}{1+\frac{x^2}{8\sqrt{2}+12}}$
$=(\frac{1}{2(8\sqrt{2}-12)}+\frac{1}{\sqrt{2}(8\sqrt{2}+12)})\int \frac{dx}{1-\frac{x^2}{4(2\sqrt{2}-3)}}$
$=\frac{1}{4}(\sqrt{2}-1)\sqrt{3+2\sqrt{2}}({\tan }^{-1}\left(\frac{x}{2\sqrt{3-2\sqrt{2}}}\right)-{\tan }^{-1}\frac{x}{\sqrt{(8\sqrt{2}+12)}})+c$
$=-\frac{1}{4}{\tan }^{-1}\frac{4x}{x^2+4}+c=\frac{1}{4}{\tan }^{-1}\frac{x^2+4}{4x}+c$