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Q.
Let $\int \frac{\left(x^{6}-4\right) d x}{\left(x^{6}+2\right)^{1 / 4} \cdot x^{4}}=\frac{l\left(x^{6}+2\right)^{m}}{x^{n}}+C$, then $\frac{n}{l m}$ is equal to
Integrals
Solution:
Let $\int \frac{\left(x^{6}-4\right) d x}{\left(x^{6}+2\right)^{1 / 4}: x^{n}}$
$=\int \frac{\left(x-\frac{4}{x^{5}}\right)}{\left(x^{2}+\frac{2}{x^{4}}\right)}$
Let $x^{2}+\frac{2}{x^{4}}=t^{4}$
$\Rightarrow\left(x-\frac{4}{x^{5}}\right) d x=2 t^{3} d t$
$\Rightarrow I=2 \int \frac{t^{3} d t}{t}=\frac{2 t^{3}}{3}=\frac{2}{3} \frac{\left(x^{\circ}+2\right)^{3 / 4}}{x^{3}}+C$
$=\frac{n}{l m}=6$