∫dx2−3x−x2 equal

I=∫dx2−3x−x2 =∫dx1722−x+322Put x+3/2=t⇒dx=dtI=∫dx1722−t2=sin−1⁡t17/2+c=sin−1⁡2x+317+c

∫dx2−3x−x2 equal

I=∫dx2−3x−x2 =∫dx1722−x+322Put x+3/2=t⇒dx=dtI=∫dx1722−t2=sin−1⁡t17/2+c=sin−1⁡2x+317+c

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$\int \frac{d x}{\sqrt{2 - 3 x - x^{2}}}$ equal

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$\tan^{- 1} (\frac{2 x + 3}{\sqrt{17}}) + c$

$\sec^{- 1} (\frac{2 x + 3}{\sqrt{17}}) + c$

$\sin^{- 1} (\frac{2 x + 3}{\sqrt{17}}) + c$

$\cos^{- 1} (\frac{2 x + 3}{\sqrt{17}}) + c$

answer is C.

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Detailed Solution

$I = \int \frac{d x}{\sqrt{2 - 3 x - x^{2}}} = \int \frac{d x}{\sqrt{{(\frac{\sqrt{17}}{2})}^{2} - {(x + \frac{3}{2})}^{2}}}$

Put $x + 3 / 2 = t \Rightarrow d x = d t$

$\begin{array}{l} I = \int \frac{d x}{\sqrt{{(\frac{\sqrt{17}}{2})}^{2} - t^{2}}} = \sin^{- 1} (\frac{t}{\sqrt{17} / 2}) + c \\ = \sin^{- 1} (\frac{2 x + 3}{\sqrt{17}}) + c \end{array}$

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$\int \frac{d x}{\sqrt{2 - 3 x - x^{2}}}$ equal