If I-8x-11-5-2x-x2dx-p-5-2x-x2-qsin-1x-1-6-C- then the value of -p-q- -p-q - is where C is integration constant

I=∫8x 11√(5+2x x^2)dx =∫ 4(2 2x) 3√(5+2x x^2)dx = 4∫(2 2x)√(5+2x x^2)dx 3 ∫dx√(6 (x 1)^2) = 8√(5+2x x^2) 3sin^ 1(x 1√(6))+C ⇒ nbsp; p= 8,q= 3 ∴ nbsp;| ...

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Byju's Answer

Standard XII

Mathematics

Odd Function

If I=∫8x-11√5...

Question

If $I = \int \frac{8 x - 11}{\sqrt{5 + 2 x - x^{2}}} d x = p \sqrt{5 + 2 x - x^{2}} + q {sin}^{- 1} (\frac{x - 1}{\sqrt{6}}) + C,$ then the value of $| p + q |$ ; $(p, q \in R)$ is
(where $C$ is integration constant)

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Solution

$I = \int \frac{8 x - 11}{\sqrt{5 + 2 x - x^{2}}} d x = \int \frac{- 4 (2 - 2 x) - 3}{\sqrt{5 + 2 x - x^{2}}} d x = - 4 \int \frac{(2 - 2 x)}{\sqrt{5 + 2 x - x^{2}}} d x - 3 \int \frac{d x}{\sqrt{6 - (x - 1)^{2}}} = - 8 \sqrt{5 + 2 x - x^{2}} - 3 {sin}^{- 1} (\frac{x - 1}{\sqrt{6}}) + C \Rightarrow p = - 8, q = - 3 ∴ | p + q | = | - 8 - 3 | = 11$

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Odd Function

Standard XII Mathematics

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If I=∫8x−11√5+2x−x2dx=p√5+2x−x2+qsin−1(x−1√6)+C, then the value of |p+q| ;(p,q∈R) is (where C is integration constant)

I=∫8x−11√5+2x−x2dx =∫−4(2−2x)−3√5+2x−x2dx =−4∫(2−2x)√5+2x−x2dx−3∫dx√6−(x−1)2 =−8√5+2x−x2−3sin−1(x−1√6)+C⇒p=−8,q=−3∴|p+q|=|−8−3|=11

If $I = \int \frac{8 x - 11}{\sqrt{5 + 2 x - x^{2}}} d x = p \sqrt{5 + 2 x - x^{2}} + q {sin}^{- 1} (\frac{x - 1}{\sqrt{6}}) + C,$ then the value of $| p + q |$ ; $(p, q \in R)$ is
(where $C$ is integration constant)