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Example 22 - Find (i) ex ( tan-1 x + 1 / 1 + x2) dx - Examples

Example 22 - Chapter 7 Class 12 Integrals - Part 2
Example 22 - Chapter 7 Class 12 Integrals - Part 3
Example 22 - Chapter 7 Class 12 Integrals - Part 4
Example 22 - Chapter 7 Class 12 Integrals - Part 5


Transcript

Example 22 Find (i) ∫1▒𝑒^π‘₯ (tan^(βˆ’1)⁑π‘₯+ 1/(1 + π‘₯^2 )) 𝑑π‘₯ ∫1▒〖𝑒^π‘₯ (tan^(βˆ’1)⁑π‘₯+1/(1 + π‘₯^2 ))𝑑π‘₯γ€— It is of the form ∫1▒〖𝑒^π‘₯ [𝑓(π‘₯)+𝑓^β€² (π‘₯)] γ€— 𝑑π‘₯=𝑒^π‘₯ 𝑓(π‘₯)+𝐢 Where 𝑓(π‘₯)=tan^(βˆ’1)⁑π‘₯ 𝑓^β€² (π‘₯)= 1/(1 + π‘₯^2 ) So, our equation becomes ∫1▒〖𝑒^π‘₯ (tan^(βˆ’1)⁑π‘₯+1/(1 + π‘₯^2 ))𝑑π‘₯γ€—=𝒆^𝒙 γ€–π­πšπ§γ€—^(βˆ’πŸ)⁑〖𝒙+π‘ͺγ€— Example 22 Find (ii) ∫1β–’((π‘₯^2 + 1) 𝑒^π‘₯)/(π‘₯ + 1)^2 𝑑π‘₯ ∫1β–’γ€–(π‘₯^2 + 1)/(π‘₯ + 1)^2 .𝑒^π‘₯ 𝑑π‘₯γ€— Adding and subtracting 1 in numerator =∫1β–’γ€–(π‘₯^2+ 1 + 1 βˆ’ 1)/(π‘₯ + 1)^2 .𝑒^π‘₯ .𝑑π‘₯γ€— =∫1β–’γ€–(π‘₯^2 βˆ’ 1 + 1 + 1)/(π‘₯ + 1)^2 .𝑒^π‘₯ .𝑑π‘₯γ€— =∫1β–’γ€–[(π‘₯^2 βˆ’ 1)/(π‘₯ + 1)^2 +2/(π‘₯ + 1)^2 ] 𝑒^π‘₯ 𝑑π‘₯γ€— =∫1▒〖𝑒^π‘₯ [(π‘₯^2 βˆ’ (1)^2)/(π‘₯ + 1)^2 +2/(π‘₯ + 1)^2 ]𝑑π‘₯γ€— =∫1▒〖𝑒^π‘₯ [(π‘₯ βˆ’ 1)(π‘₯ + 1)/(π‘₯ + 1)^2 +2/(π‘₯ + 1)^2 ]𝑑π‘₯γ€— =∫1▒〖𝑒^π‘₯ [(π‘₯ βˆ’ 1)/(π‘₯ + 1)+2/(π‘₯ + 1)^2 ]𝑑π‘₯γ€— It is of form ∫1▒〖𝑒^π‘₯ [𝑓(π‘₯)+𝑓^β€² (π‘₯)] γ€— 𝑑π‘₯=𝑒^π‘₯ 𝑓(π‘₯)+𝐢 Where 𝑓(π‘₯)=(π‘₯ βˆ’ 1)/(π‘₯ + 1) 𝑓^β€² (π‘₯)=𝑑/𝑑π‘₯ [(π‘₯ βˆ’ 1)/(π‘₯ + 1)] 𝑓^β€² (π‘₯)=(1.(π‘₯ + 1) βˆ’1 (π‘₯ βˆ’ 1))/(π‘₯ + 1)^2 =(π‘₯ + 1 βˆ’ π‘₯ + 1)/(π‘₯ + 1)^2 =2/(π‘₯ + 1)^2 Thus, our equation becomes ∫1β–’γ€–(π‘₯^2 + 1)/(π‘₯ + 1)^2 .𝑒^π‘₯=∫1▒〖𝑒^π‘₯ [(π‘₯ βˆ’ 1)/(π‘₯ + 1)+2/(π‘₯ + 1)^2 ]𝑑π‘₯γ€—γ€— =𝑒^π‘₯ [(π‘₯ βˆ’ 1)/(π‘₯ + 1)]+𝐢 =(𝒙 βˆ’ 𝟏)/(𝒙 + 𝟏).𝒆^𝒙+π‘ͺ

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.