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\int\sqrt{x^2+4}dx

Integral of (x^2+4)^0.5

Answer

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

Step-by-step explanation

Problem

$\int\sqrt{x^2+4}dx$
1

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

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

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Answer

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