Question

Evaluate: $\int \frac{x^2+4}{x^4+16}dx$

Answer 1

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

$\int \frac{1+\frac{4}{x^2}}{x^2+\frac{16}{x^2}}dx$

Let $x-\frac{4}{x}=t$ $x^2+\frac{16}{x^2}-8=t^2$

$(1+\frac{4}{x^2})dx=dt$ $x^2+\frac{16}{x^2}=t^2+8$

$\therefore$ Integration becomes

$\int \frac{dt}{t^2+8}=\int \frac{dt}{(t{)}^2+(2\sqrt{2}{)}^2}$

$=\frac{1}{2\sqrt{2}}{\tan }^{-1}\left(\frac{t}{2\sqrt{2}}\right)+C$

$=\frac{1}{2\sqrt{2}}{\tan }^{-1}\left(\frac{x-\frac{4}{x}}{2\sqrt{2}}\right)+C$

$=\frac{1}{2\sqrt{2}}{\tan }^{-1}\left(\frac{x^2-4}{2\sqrt{2}x}\right)+C$.