If int(3 x+2) sqrt{2 x^2+3 x+4} d x=f(x) sqrt | vedclass

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(B) Let $I = \int(3x+2) \sqrt{2x^2+3x+4} dx$.We express $3x+2 = \lambda(4x+3) + \mu$.Comparing coefficients: $3 = 4\lambda \Rightarrow \lambda = \frac

Vedclass · Accepted Answer

$\left(\frac{137}{32}, \frac{-23}{64 \sqrt{2}}\right)$

Vedclass · Answer

(B) Let $I = \int(3x+2) \sqrt{2x^2+3x+4} dx$.We express $3x+2 = \lambda(4x+3) + \mu$.Comparing coefficients: $3 = 4\lambda \Rightarrow \lambda = \frac{3}{4}$ and $2 = 3\lambda + \mu \Rightarrow \mu = 2 - \frac{9}{4} = -\frac{1}{4}$.Thus, $I = \frac{3}{4} \int(4x+3) \sqrt{2x^2+3x+4} dx - \frac{1}{4} \int \sqrt{2x^2+3x+4} dx$.The first part is $\frac{3}{4} \cdot \frac{2}{3} (2x^2+3x+4)^{3/2} = \frac{1}{2} (2x^2+3x+4)^{3/2}$.The second part is $-\frac{\sqrt{2}}{4} \int \sqrt{x^2 + \frac{3}{2}x + 2} dx = -\frac{\sqrt{2}}{4} \int \sqrt{(x+\frac{3}{4})^2 + \frac{23}{16}} dx$.Using $\int \sqrt{t^2+a^2} dt = \frac{t}{2}\sqrt{t^2+a^2} + \frac{a^2}{2} \sinh^{-1}(\frac{t}{a})$, we get:$I = \frac{1}{2}(2x^2+3x+4)^{3/2} - \frac{\sqrt{2}}{4} [\frac{x+3/4}{2}\sqrt{x^2+\frac{3}{2}x+2} + \frac{23/16}{2} \sinh^{-1}(\frac{x+3/4}{\sqrt{23}/4})] + C$.Simplifying, $I = \frac{1}{2}(2x^2+3x+4)^{3/2} - \frac{1}{32}(4x+3)\sqrt{2x^2+3x+4} - \frac{23}{64\sqrt{2}} \sinh^{-1}(\frac{4x+3}{\sqrt{23}}) + C$.Comparing with the given form, $f(x) = \frac{1}{2}(2x^2+3x+4) - \frac{1}{32}(4x+3)$ and $A = -\frac{23}{64\sqrt{2}}$.Calculating $f(1) = \frac{1}{2}(2+3+4) - \frac{1}{32}(4+3) = \frac{9}{2} - \frac{7}{32} = \frac{144-7}{32} = \frac{137}{32}$.Thus, $(f(1), A) = (\frac{137}{32}, -\frac{23}{64\sqrt{2}})$. The correct option is $B$.

If $\int(3 x+2) \sqrt{2 x^2+3 x+4} d x=f(x) \sqrt{2 x^2+3 x+4}+A \sinh ^{-1}\left(\frac{4 x+3}{\sqrt{23}}\right)+C$, then the ordered pair $(f(1), A)=$

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