If int frac{x^2-x+2}{x^2+x+2} d x=x-log (f(x) | vedclass

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(A) We have the integral $I = \int \frac{x^2-x+2}{x^2+x+2} dx$. First,rewrite the numerator: $x^2-x+2 = (x^2+x+2) - 2x$. So,$I = \int \left( 1 - \frac

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(A) We have the integral $I = \int \frac{x^2-x+2}{x^2+x+2} dx$. First,rewrite the numerator: $x^2-x+2 = (x^2+x+2) - 2x$. So,$I = \int \left( 1 - \frac{2x}{x^2+x+2} \right) dx = x - \int \frac{2x}{x^2+x+2} dx$. To integrate $\int \frac{2x}{x^2+x+2} dx$,write $2x = (2x+1) - 1$. Then $\int \frac{2x+1-1}{x^2+x+2} dx = \int \frac{2x+1}{x^2+x+2} dx - \int \frac{1}{(x+1/2)^2 + 7/4} dx$. This gives $\log(x^2+x+2) - \frac{1}{\sqrt{7}/2} \operatorname{Tan}^{-1}(\frac{x+1/2}{\sqrt{7}/2}) = \log(x^2+x+2) - \frac{2}{\sqrt{7}} \operatorname{Tan}^{-1}(\frac{2x+1}{\sqrt{7}})$. Substituting back,$I = x - \log(x^2+x+2) + \frac{2}{\sqrt{7}} \operatorname{Tan}^{-1}(\frac{2x+1}{\sqrt{7}}) + c$. Comparing with the given form,$f(x) = x^2+x+2$ and $g(x) = \frac{2x+1}{\sqrt{7}}$. Then $f(-1) = (-1)^2 - 1 + 2 = 2$ and $g(-1) = \frac{2(-1)+1}{\sqrt{7}} = -\frac{1}{\sqrt{7}}$. Thus,$f(-1) + \sqrt{7} g(-1) = 2 + \sqrt{7}(-\frac{1}{\sqrt{7}}) = 2 - 1 = 1$.

If $\int \frac{x^2-x+2}{x^2+x+2} d x=x-\log (f(x))+\frac{2}{\sqrt{7}} \operatorname{Tan}^{-1}(g(x))+c$,then $f(-1)+\sqrt{7} g(-1)=$

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