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

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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise
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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 )γ€— ]𝑑π‘₯γ€— Putting 𝑓(π‘₯)=tan^(βˆ’1)⁑π‘₯ 𝑓^β€² (π‘₯)= 1/(1 + π‘₯^2 ) Replacing 𝑓(π‘₯) by (tan^(βˆ’1)⁑π‘₯ ) and 𝑓^β€² (π‘₯) by 1/(1 +γ€– π‘₯γ€—^2 ), we get ∫1▒〖𝑒^π‘₯ [tan^(βˆ’1)⁑〖π‘₯+1/(1+π‘₯^2 )γ€— ]𝑑π‘₯=𝑒^π‘₯ [tan^(βˆ’1)⁑π‘₯ ]+𝐢〗 =𝑒^π‘₯ tan^(βˆ’1)⁑〖π‘₯+𝐢〗 Example 22 Find (ii) ∫1β–’((π‘₯^2 + 1) 𝑒^π‘₯)/(π‘₯ + 1)^2 𝑑π‘₯ Integrating ∫1β–’γ€–(π‘₯^2 + 1)/(π‘₯+1)^2 .𝑒^π‘₯ 𝑑π‘₯γ€— =∫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‘₯^2βˆ’1)/(π‘₯+1)^2 +2/(π‘₯+1)^2 ] 𝑒^π‘₯ 𝑑π‘₯γ€— =∫1▒𝑒π‘₯[(π‘₯^2βˆ’(1)^2)/(π‘₯+1)^2 +2/(π‘₯+1)^2 ]𝑑π‘₯ =∫1▒𝑒π‘₯[(π‘₯βˆ’1)(π‘₯+1)/(π‘₯+1)^2 +2/(π‘₯+1)^2 ]𝑑π‘₯ =∫1▒𝑒π‘₯[(π‘₯βˆ’1)(π‘₯+1)/(π‘₯+1)^2 +2/(π‘₯+1)^2 ]𝑑π‘₯ =∫1▒𝑒π‘₯[(π‘₯ βˆ’ 1)/(π‘₯ + 1)+2/(π‘₯+1)^2 ]𝑑π‘₯ Putting 𝑓(π‘₯)=(π‘₯ βˆ’ 1)/(π‘₯ + 1) ∴ 𝑓^β€² (π‘₯)=𝑑/𝑑π‘₯ [(π‘₯ βˆ’ 1)/(π‘₯ + 1)] Here, βˆ’ 𝑒(π‘₯)=π‘₯βˆ’1 𝑓^β€² (π‘₯)=(1.(π‘₯ + 1) βˆ’1 (π‘₯ βˆ’ 1))/(π‘₯ + 1)^2 𝑓^β€² (π‘₯)=(π‘₯ + 1 βˆ’ π‘₯ + 1)/(π‘₯ + 1)^2 𝑓^β€² (π‘₯)=2/(π‘₯ + 1)^2 ∴ Replacing 𝑓(π‘₯) by (π‘₯ βˆ’ 1)/(π‘₯ + 1) and 𝑓^β€² (π‘₯) by 2/(π‘₯ + 1)^2 , we get ∫1β–’γ€–(π‘₯^2 + 1)/(π‘₯ + 1)^2 .𝑒π‘₯=∫1▒𝑒π‘₯[(π‘₯ βˆ’ 1)/(π‘₯ + 1)+2/(π‘₯ + 1)^2 ]𝑑π‘₯γ€— =𝑒^π‘₯ [(π‘₯ βˆ’ 1)/(π‘₯ + 1)]+𝐢 =(π‘₯ βˆ’ 1)/(π‘₯ + 1).𝑒^π‘₯+𝐢

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