Question

Please give ans step wise

Answer 1

QUESTION :–

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$\bf \to \displaystyle \int \bf \dfrac{1}{16 + {x}^{2}}.dx = \dfrac{1}{P} { \tan }^{ - 1} \left( \dfrac{x}{Q} \right) + C \: \: then \: \: find \: \: (P+Q)$

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ANSWER :–

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GIVEN :–

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$\bf \implies\displaystyle \int \bf \dfrac{1}{16 + {x}^{2}} .dx = \dfrac{1}{P} { \tan }^{ - 1} \left( \dfrac{x}{Q} \right) + C$

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TO FIND :–

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• P + Q = ?

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SOLUTION :–

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$\bf \implies\displaystyle \int \bf \dfrac{1}{16 + {x}^{2}} .dx = \dfrac{1}{P} { \tan }^{ - 1} \left( \dfrac{x}{Q} \right) + C$

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• We should write this as –

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$\bf \implies \dfrac{1}{16} \displaystyle \int \bf \dfrac{1}{1 + \left( \dfrac{{x}^{2}}{16} \right)}.dx = \dfrac{1}{P} { \tan }^{ - 1} \left( \dfrac{x}{Q} \right) + C$

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$\bf \implies \dfrac{1}{16} \displaystyle \int \bf \dfrac{1}{1 + \left( \dfrac{{x}}{4} \right)^{2} } .dx = \dfrac{1}{P} { \tan }^{ - 1} \left( \dfrac{x}{Q} \right) + C$

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• Put x/4 = t –

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$\bf \implies \dfrac{dx}{4} = dt$

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$\bf \implies dx=4dt$

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• So that –

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$\bf \implies \dfrac{1}{16} \displaystyle \int \bf \dfrac{4.dt}{1 + \left(t \right)^{2} } = \dfrac{1}{P} { \tan }^{ - 1} \left( \dfrac{x}{Q} \right) + C$

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$\bf \implies \dfrac{1}{4} \displaystyle \int \bf \dfrac{dt}{1 + \left(t \right)^{2} } = \dfrac{1}{P} { \tan }^{ - 1} \left( \dfrac{x}{Q} \right) + C$

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• We know that –

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$\bf \implies \displaystyle \int \bf \dfrac{dx}{1 + \left(x \right)^{2} } = { \tan }^{ - 1} \left(x \right) + C$

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• So that –

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$\bf \implies \dfrac{1}{4} { \tan}^{ - 1}(t ) + C = \dfrac{1}{P} { \tan }^{ - 1} \left( \dfrac{x}{Q} \right) + C$

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• Now replace 't' –

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$\bf \implies \dfrac{1}{4} { \tan}^{ - 1}\left( \dfrac{{x}}{4} \right) + C = \dfrac{1}{P} { \tan }^{ - 1} \left( \dfrac{x}{Q} \right) + C$

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• Now compare –

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$\bf \implies \large{ \boxed{ \bf P = 4\:,\:Q = 4}}$

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▪︎ Hence –

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$\bf \implies P +Q = 4 +4$

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$\bf \implies \large{ \boxed{ \bf P +Q = 8}}$

Answer 2

Explanation:

QUESTION :–

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$\bf \to \displaystyle \int \bf \dfrac{1}{16 + {x}^{2}}.dx = \dfrac{1}{P} { \tan }^{ - 1} \left( \dfrac{x}{Q} \right) + C \: \: then \: \: find \: \: (P+Q)$

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ANSWER :–

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GIVEN :–

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$\bf \implies\displaystyle \int \bf \dfrac{1}{16 + {x}^{2}} .dx = \dfrac{1}{P} { \tan }^{ - 1} \left( \dfrac{x}{Q} \right) + C$

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TO FIND :–

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• P + Q = ?

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SOLUTION :–

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$\bf \implies\displaystyle \int \bf \dfrac{1}{16 + {x}^{2}} .dx = \dfrac{1}{P} { \tan }^{ - 1} \left( \dfrac{x}{Q} \right) + C$

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• We should write this as –

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$\bf \implies \dfrac{1}{16} \displaystyle \int \bf \dfrac{1}{1 + \left( \dfrac{{x}^{2}}{16} \right)}.dx = \dfrac{1}{P} { \tan }^{ - 1} \left( \dfrac{x}{Q} \right) + C$

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$\bf \implies \dfrac{1}{16} \displaystyle \int \bf \dfrac{1}{1 + \left( \dfrac{{x}}{4} \right)^{2} } .dx = \dfrac{1}{P} { \tan }^{ - 1} \left( \dfrac{x}{Q} \right) + C$

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• Put x/4 = t –

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$\bf \implies \dfrac{dx}{4} = dt$

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$\bf \implies dx=4dt$

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• So that –

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$\bf \implies \dfrac{1}{16} \displaystyle \int \bf \dfrac{4.dt}{1 + \left(t \right)^{2} } = \dfrac{1}{P} { \tan }^{ - 1} \left( \dfrac{x}{Q} \right) + C$

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$\bf \implies \dfrac{1}{4} \displaystyle \int \bf \dfrac{dt}{1 + \left(t \right)^{2} } = \dfrac{1}{P} { \tan }^{ - 1} \left( \dfrac{x}{Q} \right) + C$

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• We know that –

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$\bf \implies \displaystyle \int \bf \dfrac{dx}{1 + \left(x \right)^{2} } = { \tan }^{ - 1} \left(x \right) + C$

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• So that –

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$\bf \implies \dfrac{1}{4} { \tan}^{ - 1}(t ) + C = \dfrac{1}{P} { \tan }^{ - 1} \left( \dfrac{x}{Q} \right) + C$

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• Now replace 't' –

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$\bf \implies \dfrac{1}{4} { \tan}^{ - 1}\left( \dfrac{{x}}{4} \right) + C = \dfrac{1}{P} { \tan }^{ - 1} \left( \dfrac{x}{Q} \right) + C$

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• Now compare –

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$\bf \implies \large{ \boxed{ \bf P = 4\:,\:Q = 4}}$

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▪︎ Hence –

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$\bf \implies P +Q = 4 +4$

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$\bf \implies \large{ \boxed{ \bf P +Q = 8}}$