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Question Number 164376 by muneer0o0 last updated on 16/Jan/22

Answered by alephzero last updated on 16/Jan/22

$$\int \frac{\mathrm{d}x}{\sqrt{x^3+64}}=?$$$$x^3={\left(\sqrt{x^3}\right)}^2$$$$\int \frac{\mathrm{d}u}{\sqrt{u^2+a}}=\ln \mid u+\sqrt{u^2+a}\mid +\mathrm{C}$$$$\Rightarrow \int \frac{\mathrm{d}x}{\sqrt{{\left(\sqrt{x^3}\right)}^2+64}}=\ln \mid \sqrt{x^3}+\sqrt{x^3+64}\mid +\mathrm{C}$$$$\mathrm{I}\mathrm{think}\mathrm{I}\mathrm{wrong}.$$$$\left(\mathrm{I}\mathrm{just}\mathrm{started}\mathrm{learn}\mathrm{calculus}\right).$$

Commented by mr W last updated on 16/Jan/22

$$yes,verywrong!{it}^{\prime }sevennotintegrable$$$$usingelementaryfunctions.$$

Answered by ajfour last updated on 16/Jan/22

$$I=\int \frac{dx}{{(x^3+a^3)}^{1/2}}$$$$letx=a{\left(\tan {}^2\theta \right)}^{1/3}$$$$dx=\frac{2(1+\tan {}^2\theta )d\theta }{3{\left(\tan \theta \right)}^{1/3}}$$$$I=\int \frac{2d\theta }{3a\sqrt{a}{\left(\tan \theta \right)}^{1/3}\cos \theta }$$$$=\frac{2}{3a\sqrt{a}}\int \frac{d\theta }{\sin {}^{1/3}\theta \cos {}^{2/3}\theta }$$$$IthinknowGammafunction$$$$shouldbeused...$$

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