Ex 7.9, 7 - Evaluate definite integral dx / x2 + 2x + 5 - Ex 7.9 - Ex 7.9

part 2 - Ex 7.9, 7 - Ex 7.9 - Serial order wise - Chapter 7 Class 12 Integrals
part 3 - Ex 7.9, 7 - Ex 7.9 - Serial order wise - Chapter 7 Class 12 Integrals

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Ex 7.9, 7 Evaluate the integrals using substitution ∫_(āˆ’1)^(1 )▒〖 š‘‘š‘„/(š‘„^2 + 2š‘„ + 5)怗 we can write ∫_(āˆ’1)^1ā–’ć€–š‘‘š‘„/(š‘„^2 + 2š‘„ + 5)=∫_(āˆ’1)^1ā–’š‘‘š‘„/((š‘„ + 2š‘„ + 1) + 4)怗 =∫_(āˆ’1)^1ā–’š‘‘š‘„/((š‘„ + 1)^2 +怖 2怗^2 ) Putting š‘„+1=š‘” Differentiating w.r.t.š‘„ š‘‘/š‘‘š‘„ (š‘„+1)=š‘‘š‘”/š‘‘š‘„ 1=š‘‘š‘”/š‘‘š‘„ š‘‘š‘„=š‘‘š‘” Hence when š‘„ varies from – 1 to 1 then š‘” varies from 0 to 2 Therefore, ∫_(āˆ’1)^1ā–’ć€–š‘‘š‘„/((š‘„+1)^2 + 2^2 )=∫_0^2ā–’š‘‘š‘”/(š‘”^2 + 2^2 )怗 =[1/2 tan^(āˆ’1)ā”ć€–š‘”/2怗 ]_0^2 =1/2 tan^(āˆ’1)⁔〖2/2āˆ’1/2 tan^(āˆ’1)⁔〖0/2怗 怗 =1/2 tan^(āˆ’1)⁔〖1āˆ’1/2 tan^(āˆ’1)⁔0 怗 =1/2 Ɨ šœ‹/4āˆ’0 =š…/šŸ–

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