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Divide $x^4+8x^2+8$ by $x^3-4x$
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$\begin{array}{l}\phantom{\phantom{;}x^{3}-4x\phantom{;};}{\phantom{;}x\phantom{;}\phantom{-;x^n}}\\\phantom{;}x^{3}-4x\phantom{;}\overline{\smash{)}\phantom{;}x^{4}\phantom{-;x^n}+8x^{2}\phantom{-;x^n}+8\phantom{;}\phantom{;}}\\\phantom{\phantom{;}x^{3}-4x\phantom{;};}\underline{-x^{4}\phantom{-;x^n}+4x^{2}\phantom{-;x^n}\phantom{-;x^n}}\\\phantom{-x^{4}+4x^{2};}\phantom{;}12x^{2}\phantom{-;x^n}+8\phantom{;}\phantom{;}\\\end{array}$
Learn how to solve problems step by step online. Find the integral int((x^4+8x^2+8)/(x^3-4x))dx. Divide x^4+8x^2+8 by x^3-4x. Resulting polynomial. Expand the integral \int\left(x+\frac{12x^{2}+8}{x^3-4x}\right)dx into 2 integrals using the sum rule for integrals, to then solve each integral separately. The integral \int xdx results in: \frac{1}{2}x^2.