Answer
-$\frac{3}{4}$
Work Step by Step
$\int^{1}_{\sqrt{2}}(\frac{u^7}{2}-\frac{1}{u^5})du$
=$\int^{1}_{\sqrt{2}}(\frac{u^7}{2}-u^{-5})du$
=$[\frac{u^8}{16}-\frac{1}{4(u)^4}]^1_{\sqrt{2}}$
=$(\frac{1^8}{16}-\frac{1}{4(1)^4})-(\frac{\sqrt{2}^8}{16}+\frac{1}{4(\sqrt{2})^4})$
=$\frac{-3}{4}$
