For x geq 0, int sqrt{x^2+2x} , dx is equal t | vedclass

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(D) Let $I = \int \sqrt{x^2+2x} \, dx$. Completing the square, we have $x^2+2x = (x+1)^2 - 1$. So, $I = \int \sqrt{(x+1)^2 - 1} \, dx$. Using the stan

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$\frac{x+1}{2} \sqrt{x^2+2x} - \frac{1}{2} \cosh^{-1}(x+1) + C$

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(D) Let $I = \int \sqrt{x^2+2x} \, dx$. Completing the square, we have $x^2+2x = (x+1)^2 - 1$. So, $I = \int \sqrt{(x+1)^2 - 1} \, dx$. Using the standard integral formula $\int \sqrt{u^2-a^2} \, du = \frac{u}{2} \sqrt{u^2-a^2} - \frac{a^2}{2} \cosh^{-1} \left( \frac{u}{a} \right) + C$, where $u = x+1$ and $a = 1$: $I = \frac{x+1}{2} \sqrt{(x+1)^2-1} - \frac{1^2}{2} \cosh^{-1} \left( \frac{x+1}{1} \right) + C$. Thus, $I = \frac{x+1}{2} \sqrt{x^2+2x} - \frac{1}{2} \cosh^{-1}(x+1) + C$.

For $x \geq 0$, $\int \sqrt{x^2+2x} \, dx$ is equal to

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