Ex 7.2, 7 - Chapter 7 Class 12 Integrals (Term 2)
Last updated at Dec. 20, 2019 by Teachoo
Transcript
Ex 7.2, 7 Integrate the function: π₯β(π₯+2) π₯β(π₯+2) Step 1: Let (π₯+2)=π‘ Differentiating both sides π€.π.π‘.π₯ 1+0 = ππ‘/ππ₯ 1= ππ‘/ππ₯ ππ₯=ππ‘ Step 2: Integrating the function β«1βγ" " π₯β(π₯+2)γ .ππ₯ Putting the value of π₯+2 & ππ₯ . = β«1βγπ₯βπ‘γ .ππ₯ = β«1βγπ₯βπ‘γ .ππ‘ = β«1βγ(π‘β2) βπ‘γ .ππ‘ = β«1βγ(π‘β2) π‘^(1/2) γ .ππ‘ = β«1β(π‘.π‘^(1/2)β2.π‘^(1/2) ) .ππ‘ = β«1β(π‘^(3/2)β2.π‘^(1/2) ) .ππ‘ = β«1βπ‘^(3/2) .ππ‘ β 2β«1βπ‘^(1/2) .ππ‘ (Using π₯+2=π‘, π₯=π‘β2) = π‘^(3/2 + 1)/(3/2 + 1) β 2 . π‘^(1/2 + 1)/(1/2 + 1) + πΆ = π‘^(5/2)/(5/2) β 2 . π‘^(3/2)/(3/2) + πΆ = 2/5 π‘^(5/2) β 2 Γ 2/3 π‘^(3/2) + πΆ = 2/5 π‘^(5/2) β 4/3 π‘^(3/2) + πΆ Putting back π‘=π₯+2 = π/π (π+π)^(π/π) β π/π (π+π)^(π/π) + πͺ (Using β«1βπ₯^π . ππ₯=π₯^(π+1)/(π +1) )
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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.