∫dx8+3x−x2 is equal to - Infinity Learn by Sri Chaitanya

A
$\frac{2}{3} \sin^{- 1} (\frac{2 x - 1}{\sqrt{41}}) + C$
B
$\frac{3}{2} \sin^{- 1} (\frac{2 x - 3}{\sqrt{41}}) + C$
C
$\frac{1}{\sqrt{4} 1} \sin^{- 1} (\frac{2 x - 3}{\sqrt{41}}) + C$
D
$\sin^{- 1} (\frac{2 x - 3}{\sqrt{41}}) + C$

Solution:

Let

$\begin{matrix} I = \int \frac{1}{\sqrt{8 + 3 x - x^{2}}} dx = \int \frac{1}{\sqrt{8 - [x^{2} - 3 x + {(\frac{3}{2})}^{2} - {(\frac{3}{2})}^{2}]}} dx \\ = \int \frac{1}{\sqrt{8 - [{(x - \frac{3}{2})}^{2} - \frac{9}{4}]}} dx \\ = \int \frac{1}{\sqrt{8 + \frac{9}{4} - {(x - \frac{3}{2})}^{2}}} dx = \int \frac{1}{\sqrt{{(\frac{\sqrt{41}}{2})}^{2} - {(x - \frac{3}{2})}^{2}}} dx \end{matrix}$

Let $x - \frac{3}{2} = t \Rightarrow dx = dt$

$\begin{array}{l} ∴ I = \int \frac{1}{\sqrt{{(\frac{\sqrt{41}}{2})}^{2} - t^{2}}} dt = \sin^{- 1} (\frac{t}{\frac{\sqrt{41}}{2}}) + C \\ [∵ \int \frac{dx}{\sqrt{a^{2} - x^{2}}} = \sin^{- 1} (\frac{x}{a})] \\ = \sin^{- 1} (\frac{x - \frac{3}{2}}{\frac{\sqrt{41}}{2}}) + C = \sin^{- 1} (\frac{2 x - 3}{\sqrt{41}}) + C [∵ t = x - \frac{3}{2}] \end{array}$

$\int \frac{dx}{\sqrt{8 + 3 x - x^{2}}}$ is equal to

Solution:

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