Evaluate the integral: int sqrt{x^2+x+1} , dx | vedclass

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(A) To evaluate $\int \sqrt{x^2+x+1} \, dx$,we complete the square inside the square root: $x^2+x+1 = (x^2+x+\frac{1}{4}) + \frac{3}{4} = (x+\frac{1}{

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

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(A) To evaluate $\int \sqrt{x^2+x+1} \, dx$,we complete the square inside the square root: $x^2+x+1 = (x^2+x+\frac{1}{4}) + \frac{3}{4} = (x+\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2$. Let $u = x+\frac{1}{2}$,then $du = dx$. The integral becomes $\int \sqrt{u^2 + a^2} \, du$ where $a = \frac{\sqrt{3}}{2}$. Using the standard formula $\int \sqrt{u^2+a^2} \, du = \frac{u}{2}\sqrt{u^2+a^2} + \frac{a^2}{2}\ln|u+\sqrt{u^2+a^2}| + c$. Substituting back $u = x+\frac{1}{2}$ and $a^2 = \frac{3}{4}$: $= \frac{x+1/2}{2}\sqrt{x^2+x+1} + \frac{3/4}{2}\ln|x+\frac{1}{2} + \sqrt{x^2+x+1}| + c$. $= \frac{2x+1}{4}\sqrt{x^2+x+1} + \frac{3}{8}\sinh^{-1}\left(\frac{2x+1}{\sqrt{3}}\right) + c$.

Evaluate the integral: $\int \sqrt{x^2+x+1} \, dx$

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