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Question: Answered & Verified by Expert
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 $(f(\mathbf{l}), A)=$
MathematicsIndefinite IntegrationTS EAMCETTS EAMCET 2020 (09 Sep Shift 2)
Options:
  • A $\left(\frac{73}{8}, \frac{23}{64 \sqrt{2}}\right)$
  • B $\left(\frac{137}{32}, \frac{-23}{64 \sqrt{2}}\right)$
  • C $\left(\frac{15}{8}, \frac{-23}{16 \sqrt{2}}\right)$
  • D $\left(\frac{49}{32}, \frac{23}{16 \sqrt{2}}\right)$
Solution:
1438 Upvotes Verified Answer
The correct answer is: $\left(\frac{137}{32}, \frac{-23}{64 \sqrt{2}}\right)$
Let $I=\int(3 x+2) \sqrt{2 x^2+3 x+4} d x$
$\begin{aligned} & \text { Put, } 3 x+2=\lambda(4 x+3)+\mu \\ & \Rightarrow \quad 3=4 \lambda \text { and } 2=3 \lambda+\mu\end{aligned}$
$\therefore \quad \lambda=\frac{3}{4}, \mu=\frac{-1}{4}$
$\therefore \quad I=\frac{3}{4} \int(4 x+3) \sqrt{2 x^2+3 x+4} d x$
$-\frac{1}{4} \int \sqrt{2 x^2+3 x+4} d x$
$\begin{aligned} & =\frac{1}{2}\left(2 x^2+3 x+4\right)^{3 / 2}-\frac{1}{4} \times \sqrt{2} \int \sqrt{x^2+\frac{3}{2} x+2} d x \\ & =\frac{1}{2}\left(2 x^2+3 x+4\right)^{3 / 2}-\frac{\sqrt{2}}{4} \int \sqrt{\left(x+\frac{3}{4}\right)^2+2-\frac{9}{16}} d x \\ & =\frac{1}{2}\left(2 x^2+3 x+4\right)^{3 / 2}-\frac{\sqrt{2}}{4} \\ & {\left[\frac{x+\frac{3}{4}}{2} \sqrt{\left(x+\frac{3}{4}\right)^2+\frac{23}{16}}\right]+C} \\ & =\frac{1}{2}\left(2 x^2+3 x+4\right)^{3 / 2}-\frac{\sqrt{2}}{4} \\ & {\left[\frac{4 x+3}{8} \sqrt{x^2+\frac{3}{2} x+2}\left[\left(x+\frac{3}{4}\right)+\sqrt{\left(x+\frac{3}{4}\right)^2+\frac{23}{16}}\right]\right]} \\ & \left.\quad+\frac{23}{16} \log \left[\frac{4 x+3}{4}+\sqrt{x^2+\frac{3}{2} x+\frac{32}{16}}\right]\right]+C\end{aligned}$
$=\frac{1}{2}\left(2 x^2+3 x+4\right)^{3 / 2}-\frac{\sqrt{2}}{4}$
$\begin{aligned} & {\left[\begin{array}{l}\frac{4 x+3}{8} \sqrt{x^2+\frac{3}{2} x+2} \\ +\frac{23}{16} \log \left[\frac{4 x+3}{4}+\sqrt{x^2+\frac{3}{2} x+\frac{32}{16}}\right]\end{array}\right]+C} \\ & =\frac{1}{2}\left(2 x^2+3 x+4\right)^{3 / 2}-\frac{1}{32}(4 x+3) \sqrt{2 x^2+3 x+4} \\ & \frac{-23 \sqrt{2}}{64} \cdot 2 \sinh \mathrm{h}^{-1}\left(\frac{4 x+3}{\sqrt{23}}\right)+C \\ & =\left[\frac{1}{2}\left(2 x^2+3 x+4\right)-\frac{1}{32}(4 x+3)\right] \sqrt{2 x^2+3 x+4} \\ & -\frac{23}{64 \sqrt{2}} \sinh ^{-1}\left(\frac{4 x+3}{\sqrt{23}}\right)+C \\ & \end{aligned}$
$\therefore \quad f(x)=\frac{1}{2}\left(2 x^2+3 x+4\right)-\frac{1}{32}(4 x+3)$
$\therefore \quad f(1)=\frac{9}{2}-\frac{7}{32}=\frac{137}{32} \quad$ and $A=\frac{-23}{64 \sqrt{2}}$

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