Question ID: #424
If $\int \frac{2x^{2}+5x+9}{\sqrt{x^{2}+x+1}}dx=x\sqrt{x^{2}+x+1}+\alpha\sqrt{x^{2}+x+1}+\beta \log_{e}|x+\frac{1}{2}+\sqrt{x^{2}+x+1}|+C,$ where C is the constant of integration, then $\alpha+2\beta$ is equal to:
Solution:
We express the numerator in terms of the quadratic term in the denominator and its derivative:
$$ 2x^2+5x+9 = A(x^2+x+1) + B(2x+1) + C $$
Comparing coefficients:
$A = 2$
$B = \frac{3}{2}$
$C = \frac{11}{2}$
The integral becomes:
$$ 2\int \sqrt{x^2+x+1} dx + \frac{3}{2}\int \frac{2x+1}{\sqrt{x^2+x+1}} dx + \frac{11}{2}\int \frac{dx}{\sqrt{x^2+x+1}} $$
Using standard integration formulas:
$$ 2 \left[ \frac{2x+1}{4}\sqrt{x^2+x+1} + \frac{3}{8}\ln\left(x+\frac{1}{2}+\sqrt{x^2+x+1}\right) \right] + 3\sqrt{x^2+x+1} + \frac{11}{2}\ln\left(x+\frac{1}{2}+\sqrt{x^2+x+1}\right) + C $$
Simplifying the term with $\sqrt{x^2+x+1}$:
$$ \left( \frac{2x+1}{2} + 3 \right)\sqrt{x^2+x+1} = \left( x + \frac{1}{2} + 3 \right)\sqrt{x^2+x+1} = \left( x + \frac{7}{2} \right)\sqrt{x^2+x+1} $$
So, the term $x\sqrt{x^2+x+1}$ is separated, leaving $\alpha = \frac{7}{2}$.
Combining the logarithmic terms:
$$ \left( 2 \cdot \frac{3}{8} + \frac{11}{2} \right) \ln(\dots) = \left( \frac{3}{4} + \frac{22}{4} \right) \ln(\dots) = \frac{25}{4} \ln(\dots) $$
So, $\beta = \frac{25}{4}$.
Finally, we find $\alpha + 2\beta$:
$$ \frac{7}{2} + 2\left(\frac{25}{4}\right) = \frac{7}{2} + \frac{25}{2} = \frac{32}{2} = 16 $$
Ans. (16)
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