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Question Number 111206 by mohammad17 last updated on 02/Sep/20

$$\int \frac{dx}{x^3+3x-5}$$

Answered by Sarah85 last updated on 02/Sep/20

$$\mathrm{we}\mathrm{had}\mathrm{this}\mathrm{before}$$$$\mathrm{if}\mathrm{we}\mathrm{cannot}\mathrm{find}‘‘{\mathrm{nice}}^{″}\mathrm{factors}\mathrm{of}\mathrm{the}$$$$\mathrm{denominator}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{getting}\mathrm{complicated}$$$$\mathrm{I}\mathrm{show}\mathrm{you}\mathrm{how}\mathrm{to}\mathrm{solve}$$$$\int \frac{dx}{x^3+3x-\frac{26\sqrt{3}}{9}}\mathrm{instead}.\frac{26\sqrt{3}}{9}=5.003702...$$$$\mathrm{but}{x}^3+3x-\frac{26\sqrt{3}}{9}=(x-\frac{2\sqrt{3}}{3})(x^2+\frac{2\sqrt{3}}{3}x+\frac{13}{3})$$$$\Rightarrow$$$$\int \frac{dx}{x^3+3x-\frac{26\sqrt{3}}{9}}=\int \frac{dx}{7(x-\frac{2\sqrt{3}}{3})}-\int \frac{x+\frac{4\sqrt{3}}{3}}{7(x^2+\frac{2\sqrt{3}}{3}x+\frac{13}{3})}dx$$$$\mathrm{and}\mathrm{we}\mathrm{can}‘‘{\mathrm{easily}}^{″}\mathrm{solve}\mathrm{these}$$$$=\frac{1}{7}\ln (3x-2\sqrt{3})-\frac{1}{14}\ln (3x^2+2\sqrt{3}x+12)+\frac{\sqrt{3}}{14}\arctan \frac{3x+\sqrt{3}}{6}+C$$$$\mathrm{the}\mathrm{problem}\mathrm{with}$$$$x^3+3x+5\mathrm{is}\mathrm{that}\mathrm{the}\mathrm{real}\mathrm{root}\mathrm{is}$$$$\sqrt[3]{\frac{5}{2}+\frac{\sqrt{29}}{2}}-\sqrt[3]{-\frac{5}{2}+\frac{\sqrt{29}}{2}}$$$$\mathrm{and}\mathrm{you}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{want}\mathrm{to}\mathrm{write}\mathrm{it}\mathrm{out}$$

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