If int frac{2 x^2+5 x+9}{sqrt{x^2+x+1}} d x=x | vedclass

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Question

(A) Let $2 x^2+5 x+9 = A(x^2+x+1) + B(2x+1) + D$.Comparing coefficients: $A=2$,$A+2B=5 \implies 2+2B=5 \implies B=\frac{3}{2}$,and $A+B+D=9 \implies 2

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(A) Let $2 x^2+5 x+9 = A(x^2+x+1) + B(2x+1) + D$.Comparing coefficients: $A=2$,$A+2B=5 \implies 2+2B=5 \implies B=\frac{3}{2}$,and $A+B+D=9 \implies 2+\frac{3}{2}+D=9 \implies D=\frac{11}{2}$.The integral becomes $\int \frac{2(x^2+x+1) + \frac{3}{2}(2x+1) + \frac{11}{2}}{\sqrt{x^2+x+1}} dx = 2\int \sqrt{x^2+x+1} dx + \frac{3}{2}\int \frac{2x+1}{\sqrt{x^2+x+1}} dx + \frac{11}{2}\int \frac{dx}{\sqrt{(x+\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2}}$.Using the formula $\int \sqrt{x^2+a^2} dx = \frac{x}{2}\sqrt{x^2+a^2} + \frac{a^2}{2}\ln|x+\sqrt{x^2+a^2}|$:$2[\frac{x+1/2}{2}\sqrt{x^2+x+1} + \frac{3/4}{2}\ln|x+1/2+\sqrt{x^2+x+1}|] + 3\sqrt{x^2+x+1} + \frac{11}{2}\ln|x+1/2+\sqrt{x^2+x+1}| + C$.$= (x+\frac{1}{2})\sqrt{x^2+x+1} + \frac{3}{4}\ln|...| + 3\sqrt{x^2+x+1} + \frac{11}{2}\ln|...| + C$.$= x\sqrt{x^2+x+1} + (\frac{1}{2}+3)\sqrt{x^2+x+1} + (\frac{3}{4}+\frac{11}{2})\ln|x+\frac{1}{2}+\sqrt{x^2+x+1}| + C$.Comparing with the given form: $\alpha = \frac{7}{2}$ and $\beta = \frac{3}{4} + \frac{22}{4} = \frac{25}{4}$.$\alpha + 2\beta = \frac{7}{2} + 2(\frac{25}{4}) = \frac{7}{2} + \frac{25}{2} = \frac{32}{2} = 16$.

If $\int \frac{2 x^2+5 x+9}{\sqrt{x^2+x+1}} d x=x \sqrt{x^2+x+1}+\alpha \sqrt{x^2+x+1}+\beta \log _{ e }\left| x +\frac{1}{2}+\sqrt{ x ^2+ x +1}\right|+ C$,where $C$ is the constant of integration,then $\alpha+2 \beta$ is equal to . . . . . .

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