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Step-by-step Solution

Integrate $\sqrt{x^2+4}$ with respect to x

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Answer

$\frac{x\sqrt{x^2+4}}{2}+2\ln\left|\frac{\sqrt{x^2+4}+x}{2}\right|+C_0$

Step-by-step explanation

Problem to solve:

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

Solve the integral $\int\sqrt{x^2+4}dx$ 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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Substituting in the original integral, we get

$\int2\sqrt{4\tan\left(\theta\right)^2+4}\sec\left(\theta\right)^2d\theta$

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Answer

$\frac{x\sqrt{x^2+4}}{2}+2\ln\left|\frac{\sqrt{x^2+4}+x}{2}\right|+C_0$