Calculus II - Arc Length Just wanted to verify my work and see if anything could be done more easily. Find the exact arc length of the portion of the graph of \[y = (1/6)x^3 + 1/(2x) \] from x = 1 to x=2 dy/dx = \[(x^2/2) - (1/(2x^2))\] \[(dy/dx)^2 = x^4/4 + (1)/(4x^4) - 1/2 \] \[(dy/dx)^2+1 = x^4/4 + (1)/(4x^4) + 1/2 \] \[\int\limits_{1}^{2}\sqrt{x^4/4 + 1/(4x^4) + 1/2}\] Arc Length = 17/12

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Calculus II - Arc Length Just wanted to verify my work and see if anything could be done more easily. Find the exact arc length of the portion of the graph of \[y = (1/6)x^3 + 1/(2x) \] from x = 1 to x=2 dy/dx = \[(x^2/2) - (1/(2x^2))\] \[(dy/dx)^2 = x^4/4 + (1)/(4x^4) - 1/2 \] \[(dy/dx)^2+1 = x^4/4 + (1)/(4x^4) + 1/2 \] \[\int\limits_{1}^{2}\sqrt{x^4/4 + 1/(4x^4) + 1/2}\] Arc Length = 17/12

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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My biggest question is 1) Whether the answer is right and 2) Whether it is possible to get a perfect square from (dy/dx)^2 + 1, since the integral of that was an absolute nightmare.

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