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\int\frac{x2}{\sqrt{x^2-4}}dx

Integral of (x^2)/((x^2-4)^0.5)

Answer

$4\left(\frac{1}{2}\left(\ln\left(\sqrt{x^2-4}+x\right)-\frac{4}{\sqrt{3}}\right)+\frac{\frac{1}{8}\sqrt{x^2-4}x^{2}}{x}\right)+C_0$

Step-by-step explanation

Problem

$\int\frac{x2}{\sqrt{x^2-4}}dx$
1

Solve the integral $\int\frac{x^2}{\sqrt{x^2-4}}dx$ by trigonometric substitution using the substitution

$\begin{matrix}x=2\sec\left(\theta\right) \\ dx=2\tan\left(\theta\right)\sec\left(\theta\right)d\theta\end{matrix}$

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Answer

$4\left(\frac{1}{2}\left(\ln\left(\sqrt{x^2-4}+x\right)-\frac{4}{\sqrt{3}}\right)+\frac{\frac{1}{8}\sqrt{x^2-4}x^{2}}{x}\right)+C_0$

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