Question

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

Answer 1

Let, $I=\int \frac{dx}{x^4+16}=\frac{1}{8}\int \frac{8dx}{x^4+16}=\frac{1}{8}\int \frac{\frac{8}{x^2}}{x^2+\frac{16}{x^2}}dx=\frac{1}{8}\int \frac{(1+\frac{4}{x^2})-(1-\frac{4}{x^2})}{x^2+\frac{16}{x^2}}=\frac{1}{8}\int \frac{(1+\frac{4}{x^2})}{x^2+\frac{16}{x^2}-8+8}dx-\frac{1}{8}\int \frac{(1-\frac{4}{x^2})}{x^2+\frac{16}{x^2}+8-8}dx=\frac{1}{8}\int \frac{(1+\frac{4}{x^2})}{{(x-\frac{4}{x})}^2+{\left(2\sqrt{2}\right)}^2}dx-\frac{1}{8}\int \frac{(1-\frac{4}{x^2})}{{(x+\frac{4}{x})}^2-{\left(2\sqrt{2}\right)}^2}dx$

Let, $I_1=\frac{1}{8}\int \frac{(1+\frac{4}{x^2})}{{(x-\frac{4}{x})}^2+{\left(2\sqrt{2}\right)}^2}dx$

Putting, $x-\frac{4}{x}=u\Rightarrow (1+\frac{4}{x^2})dx=du$

Now, $I_1=\frac{1}{8}\int \frac{du}{u^2+{\left(2\sqrt{2}\right)}^2}=\frac{1}{8}\times \frac{1}{2\sqrt{2}}{\tan }^{-1}\left[\frac{u}{2\sqrt{2}}\right]+C_1=\frac{1}{16\sqrt{2}}{\tan }^{-1}\left[\frac{x^2-4}{2\sqrt{2}x}\right]+C_1$

Again, let $I_2=\frac{1}{8}\int \frac{(1-\frac{4}{x^2})}{{(x+\frac{4}{x})}^2-{\left(2\sqrt{2}\right)}^2}dx$

Putting, $x+\frac{4}{x}=v\Rightarrow {(1-\frac{4}{x^2})}^{2}dx=dv$

So, $I_2=\frac{1}{8}\int \frac{dv}{v^2-{\left(2\sqrt{2}\right)}^2}=\frac{1}{8}\times \frac{1}{4\sqrt{2}}\log \left|\frac{v-2\sqrt{2}}{v+2\sqrt{2}}\right|+C_2=\frac{1}{32\sqrt{2}}\log \left|\frac{x^2+4-2\sqrt{2}x}{x^2+4+2\sqrt{2}x}\right|+C_2$

$\therefore I=\int \frac{dx}{x^4+16}=(\frac{1}{16\sqrt{2}}{\tan }^{-1}\left[\frac{x^2-4}{2\sqrt{2}x}\right]+C_1)-(\frac{1}{32\sqrt{2}}\log \left|\frac{x^2+4-2\sqrt{2}x}{x^2+4+2\sqrt{2}x}\right|+C_2)=\frac{1}{16\sqrt{2}}{\tan }^{-1}\left[\frac{x^2-4}{2\sqrt{2}x}\right]-\frac{1}{32\sqrt{2}}\log \left|\frac{x^2+4-2\sqrt{2}x}{x^2+4+2\sqrt{2}x}\right|+C.$