-2x -5-7 -6x - x2 dx - A -7 - 6x -x2 - B sin-1x - 34 - C where C is a constant of integration- then the value of B-A is equal to

7 6x x^2 = 16 (x + 3)^2 and ddx (7 6x x^2) = 2x 6 So, ∫2x +5√(7 6x x^2) dx =∫2x + 6√(7 6x x^2)dx ∫1√(16 (x +3)^2)dx = 2 √(7 6x x^2 ...

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Byju's Answer

Standard XII

Physics

Integration

∫2x +5√7 -6x ...

Question

$\int \frac{2 x + 5}{\sqrt{7 - 6 x - x^{2}}} d x = A \sqrt{7 - 6 x - x^{2}} + B {sin}^{- 1} (\frac{x + 3}{4}) + C$ (where C is a constant of integration), then the value of $B - A$ is equal to

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Solution

$7 - 6 x - x^{2} = 16 - (x + 3)^{2}$
and $\frac{d}{d x} (7 - 6 x - x^{2}) = - 2 x - 6$
So,
$\int \frac{2 x + 5}{\sqrt{7 - 6 x - x^{2}}} d x = \int \frac{2 x + 6}{\sqrt{7 - 6 x - x^{2}}} d x - \int \frac{1}{\sqrt{16 - (x + 3)^{2}}} d x$
$= - 2 \sqrt{7 - 6 x - x^{2}} - {sin}^{- 1} (\frac{x + 3}{4}) + C$
Therefore, $A = - 2, B = - 1$ , so
$B - A = 1$

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∫2x+5√7−6x−x2dx=A√7−6x−x2+Bsin−1(x+34)+C (where C is a constant of integration), then the value of B−A is equal to

7−6x−x2=16−(x+3)2 and ddx(7−6x−x2)=−2x−6 So, ∫2x+5√7−6x−x2dx=∫2x+6√7−6x−x2dx−∫1√16−(x+3)2dx =−2√7−6x−x2−sin−1(x+34)+C Therefore, A=−2,B=−1, so B−A=1

$\int \frac{2 x + 5}{\sqrt{7 - 6 x - x^{2}}} d x = A \sqrt{7 - 6 x - x^{2}} + B {sin}^{- 1} (\frac{x + 3}{4}) + C$ (where C is a constant of integration), then the value of $B - A$ is equal to