Question

Let $\int \frac{\left(x^6-4\right)dx}{{\left(x^6+2\right)}^{1/4}\cdot x^4}=\frac{l{\left(x^6+2\right)}^m}{x^n}+C$, then $\frac{n}{lm}$ is equal to

Answer 1

Answer: 6

Let $\int \frac{\left(x^6-4\right)dx}{{\left(x^6+2\right)}^{1/4}:x^n}$
$=\int \frac{\left(x-\frac{4}{x^5}\right)}{\left(x^2+\frac{2}{x^4}\right)}$
Let $x^2+\frac{2}{x^4}=t^4$
$\Rightarrow \left(x-\frac{4}{x^5}\right)dx=2t^{3}dt$
$\Rightarrow I=2\int \frac{t^{3}dt}{t}=\frac{2t^3}{3}=\frac{2}{3}\frac{{\left(x^{\circ }+2\right)}^{3/4}}{x^3}+C$
$=\frac{n}{lm}=6$