Miscellaneous

Chapter 7 Class 12 Integrals
Serial order wise

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### 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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.