I think that from this conclusion I suppose I have done something wrong at algebraic modification or I have chosen the worst way around this integral, can someone help me
2026-04-04 03:55:51.1775274951
Evaluate the integral $ \int\frac{\sqrt{x^2+\frac 1{x^2}-2}}{x^5}\sqrt[3]{x\sqrt{x}}\:dx$ without substitution
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Hint. You may observe that $$ x^2+\frac 1{x^2}-2=\left(x-\frac 1x\right)^2 $$ then $$ \frac{\sqrt{x^2+\frac 1{x^2}-2}}{x^5}\sqrt[3]{x\sqrt{x}}=\left|x-\frac 1x\right|x^{- 9/2}$$ and the evaluation is straightforward on appropriate intervals.
For example, for $x \geq 1$, you have $$ \int\frac{\sqrt{x^2+\frac 1{x^2}-2}}{x^5}\sqrt[3]{x\sqrt{x}}\:dx=\int\left(x-\frac 1x\right)x^{-9/2}dx=\int x^{-7/2}dx-\int x^{-11/2}dx$$