Misc 2 - Chapter 7 Class 12 Integrals

Last updated at April 16, 2024 by

Misc 2 - Integrate 1/root x+a + root x+b - Class 12 - Miscellaneous

Misc 2 - Chapter 7 Class 12 Integrals - Part 2
Misc 2 - Chapter 7 Class 12 Integrals - Part 3

Transcript

Misc 2 Integrate the function 1/(√(𝑥 + 𝑎) + √(𝑥 +𝑏) ) ∫1▒〖1/(√(𝑥 + 𝑎) + √(𝑥 + 𝑏) ) 𝑑𝑥〗 Rationalising = ∫1▒〖(1/(√(𝑥 + 𝑎) + √(𝑥 + 𝑏) ) × (√(𝑥 + 𝑎) − √(𝑥 + 𝑏))/(√(𝑥 + 𝑎) − √(𝑥 + 𝑏) )) 𝑑𝑥〗 = ∫1▒〖(√(𝑥 + 𝑎) − √(𝑥 + 𝑏))/((√(𝑥 + 𝑎) + √(𝑥 + 𝑏) )(√(𝑥 + 𝑎) − √(𝑥 + 𝑏) ) ) 𝑑𝑥〗 Using (𝑎−𝑏)(𝑎+𝑏)=𝑎^2−𝑏^2 =∫1▒〖(√(𝑥 + 𝑎) − √(𝑥 + 𝑏))/((√(𝑥 + 𝑎) )^2 − (√(𝑥 + 𝑏) )^2 ) 𝑑𝑥〗 = ∫1▒〖(√(𝑥 + 𝑎) − √(𝑥 + 𝑏))/(𝑥 + 𝑎 −(𝑥 + 𝑏) ) 𝑑𝑥〗 = ∫1▒〖(√(𝑥 + 𝑎) − √(𝑥 + 𝑏))/(𝑎 − 𝑏) 𝑑𝑥〗 = 1/(𝑎 − 𝑏) ∫1▒(√(𝑥+𝑎) −√(𝑥+𝑏) ) 𝑑𝑥 = 1/(𝑎 − 𝑏) ∫1▒((𝑥+𝑎)^(1/2) −(𝑥+𝑎)^(1/2) ) 𝑑𝑥 = 1/(𝑎 − 𝑏) [∫1▒(𝑥+𝑎)^(1/2) 𝑑𝑥−∫1▒(𝑥+𝑎)^(1/2) 𝑑𝑥] = 1/(𝑎 − 𝑏) [(𝑥 + 𝑎)^(1/2 + 1)/(1/2 + 1) − (𝑥 + 𝑏)^(1/2 + 1)/(1/2 + 1) ] + 𝐶 = 1/(𝑎 − 𝑏) [(𝑥 + 𝑎)^(3/2)/(3/2) − (𝑥 + 𝑏)^(3/2)/(3/2)] + 𝐶 = 1/(𝑎 − 𝑏) [〖2(𝑥 + 𝑎)〗^(3/2)/3 − 〖2(𝑥 + 𝑏)〗^(3/2)/3] + 𝐶 = 𝟐/𝟑(𝒂 − 𝒃) [(𝒙+𝒂)^(𝟑/𝟐)−(𝒙+𝒃)^(𝟑/𝟐) ] + 𝑪

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.