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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.