int sqrt{x^2+x+1} , dx times int frac{1}{sqrt | vedclass

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(B) First, we complete the square for the quadratic expression: $x^2+x+1 = (x+\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2$.The first integral is $\int \sq

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

Vedclass · Answer

(B) First, we complete the square for the quadratic expression: $x^2+x+1 = (x+\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2$.The first integral is $\int \sqrt{(x+\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} \, dx = \frac{x+\frac{1}{2}}{2} \sqrt{x^2+x+1} + \frac{3}{8} \sinh^{-1}\left(\frac{2x+1}{\sqrt{3}}\right) + C_1$.The second integral is $\int \frac{1}{\sqrt{(x+\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2}} \, dx = \sinh^{-1}\left(\frac{2x+1}{\sqrt{3}}\right) + C_2$.Multiplying these two results, we get the product as $\left(\frac{2x+1}{4} \sqrt{x^2+x+1} + \frac{3}{8} \sinh^{-1} \frac{2x+1}{\sqrt{3}}\right) \sinh^{-1}\left(\frac{2x+1}{\sqrt{3}}\right) + C$.

$\int \sqrt{x^2+x+1} \, dx \times \int \frac{1}{\sqrt{x^2+x+1}} \, dx$ is equal to

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