int frac{1}{(x+2) sqrt{x^2+x+2}} , dx = | vedclass

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(B) To solve the integral $I = \int \frac{1}{(x+2) \sqrt{x^2+x+2}} \, dx$, we use the substitution $x+2 = \frac{1}{t}$.Then $dx = -\frac{1}{t^2} \, dt

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$-\frac{1}{\sqrt{2}} \operatorname{Sinh}^{-1} \left( \frac{x+6}{\sqrt{7}(x+2)} \right) + c$

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(B) To solve the integral $I = \int \frac{1}{(x+2) \sqrt{x^2+x+2}} \, dx$, we use the substitution $x+2 = \frac{1}{t}$.Then $dx = -\frac{1}{t^2} \, dt$.Substituting these into the integral:$I = \int \frac{1}{(1/t) \sqrt{(1/t - 2)^2 + (1/t - 2) + 2}} \cdot (-1/t^2) \, dt$$I = -\int \frac{1}{t \sqrt{1/t^2 - 4/t + 4 + 1/t - 2 + 2}} \, dt$$I = -\int \frac{1}{\sqrt{1 - 3t + 4t^2}} \, dt$$I = -\frac{1}{2} \int \frac{1}{\sqrt{t^2 - \frac{3}{4}t + \frac{1}{4}}} \, dt$Completing the square: $t^2 - \frac{3}{4}t + \frac{1}{4} = (t - \frac{3}{8})^2 + \frac{1}{4} - \frac{9}{64} = (t - \frac{3}{8})^2 + \frac{7}{64}$.Using the formula $\int \frac{dx}{\sqrt{x^2+a^2}} = \ln|x + \sqrt{x^2+a^2}|$, we get:$I = -\frac{1}{2} \ln \left| (t - \frac{3}{8}) + \sqrt{(t - \frac{3}{8})^2 + \frac{7}{64}} \right| + c$.Substituting $t = \frac{1}{x+2}$ back, we obtain the result in terms of $\operatorname{Sinh}^{-1}$.

$\int \frac{1}{(x+2) \sqrt{x^2+x+2}} \, dx =$

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