∫1 / (x²(x⁴ + 2)^(3/4)) dx

Explanation:

Step1: Rewrite the integrand

First, rewrite the integrand $\frac{1}{x^{2}(x^{4}+2)^{\frac{3}{4}}}$ as $\frac{1}{x^{2}x^{3}(1 + \frac{2}{x^{4}})^{\frac{3}{4}}}=\frac{1}{x^{5}(1+\frac{2}{x^{4}})^{\frac{3}{4}}}$ by factoring out $x^{4}$ from the denominator.

Step2: Use substitution

Let $u = 1+\frac{2}{x^{4}}$, then $du=-\frac{8}{x^{5}}dx$, and $\frac{1}{x^{5}}dx=-\frac{1}{8}du$.

Step3: Integrate with respect to u

The integral $\int\frac{1}{x^{5}(1 + \frac{2}{x^{4}})^{\frac{3}{4}}}dx$ becomes $-\frac{1}{8}\int u^{-\frac{3}{4}}du$. Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $-\frac{1}{8}\times\frac{u^{-\frac{3}{4}+1}}{-\frac{3}{4}+1}+C$.

Step4: Simplify the result

$-\frac{1}{8}\times\frac{u^{\frac{1}{4}}}{\frac{1}{4}}+C=-\frac{1}{2}u^{\frac{1}{4}}+C$.

Step5: Substitute back u

Substitute $u = 1+\frac{2}{x^{4}}$ back into the result, we get $-\frac{1}{2}(1+\frac{2}{x^{4}})^{\frac{1}{4}}+C$.

Answer:

$-\frac{1}{2}(1+\frac{2}{x^{4}})^{\frac{1}{4}}+C$