Chapter 7 Class 12 Integrals

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Ex 7.4, 22 - Chapter 7 Class 12 Integrals (Important Question)

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Ex 7.4, 22 - Integrate x + 3 / x^2 - 2x - 5 - Class 12 NCERT

Ex 7.4, 22 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.4, 22 - Chapter 7 Class 12 Integrals - Part 3 Ex 7.4, 22 - Chapter 7 Class 12 Integrals - Part 4 Ex 7.4, 22 - Chapter 7 Class 12 Integrals - Part 5

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Ex 7.4, 22 Integrate the function (š‘„ + 3)/(š‘„^2 āˆ’ 2š‘„ āˆ’ 5) āˆ«1ā–’(š‘„ + 3)/(š‘„^2 āˆ’ 2š‘„ āˆ’ 5) š‘‘š‘„ =1/2 āˆ«1ā–’(2š‘„ + 6)/(š‘„^2 āˆ’ 2š‘„ āˆ’ 5) š‘‘š‘„ =1/2 āˆ«1ā–’(2š‘„ āˆ’ 2 + 2 + 6 )/(š‘„^2 āˆ’ 2š‘„ āˆ’ 5) š‘‘š‘„ =1/2 āˆ«1ā–’(2š‘„ āˆ’ 2)/(š‘„^2 āˆ’ 2š‘„ āˆ’ 5) š‘‘š‘„+8/2 āˆ«1ā–’š‘‘š‘„/(š‘„^2 āˆ’ 2š‘„ āˆ’ 5) =1/2 āˆ«1ā–’(2š‘„ āˆ’ 2)/(š‘„^2 āˆ’ 2š‘„ āˆ’ 5) š‘‘š‘„+4āˆ«1ā–’š‘‘š‘„/(š‘„^2 āˆ’ 2š‘„ āˆ’ 5) Rough (š‘„^2āˆ’2š‘„āˆ’5)^ā€²=2š‘„āˆ’2 Solving š‘°šŸ I1=1/2 āˆ«1ā–’(2š‘„ āˆ’ 2)/(š‘„^2 āˆ’ 2š‘„ āˆ’ 5) . š‘‘š‘„ Let š‘„^2 āˆ’ 2š‘„ āˆ’ 5=š‘” Diff both sides w.r.t.x 2š‘„āˆ’2āˆ’0=š‘‘š‘”/š‘‘š‘„ š‘‘š‘„=š‘‘š‘”/(2š‘„ āˆ’ 2) Thus, our equation becomes āˆ“ I1=1/2 āˆ«1ā–’(2š‘„ āˆ’ 2)/(š‘„^2 āˆ’ 2š‘„ āˆ’ 5) . š‘‘š‘„ Putting value of (š‘„^2āˆ’2š‘„āˆ’5)=š‘” and š‘‘š‘„=š‘‘š‘”/(2š‘„ āˆ’ 2) I1=1/2 āˆ«1ā–’(2š‘„ āˆ’ 2)/š‘” . š‘‘š‘„ I1=1/2 āˆ«1ā–’(2š‘„ āˆ’ 2)/š‘” . š‘‘š‘”/(2š‘„ āˆ’ 2) I1=1/2 āˆ«1ā–’1/š‘” . š‘‘š‘” I1=1/2 logā”|š‘”|+š¶1 I1=1/2 logā”|š‘„^2āˆ’2š‘„āˆ’5|+š¶1 Solving š‘°šŸ I2=4āˆ«1ā–’1/(š‘„^2 āˆ’ 2š‘„ āˆ’ 5) . š‘‘š‘„ (Using š‘”=š‘„^2āˆ’2š‘„āˆ’5) I2=4āˆ«1ā–’1/(š‘„^2 āˆ’ 2(š‘„)(1) āˆ’ 5) . š‘‘š‘„ I2=4āˆ«1ā–’1/(š‘„^2 āˆ’ 2(š‘„)(1) + (1)^2 āˆ’ (1)^2 āˆ’ 5) . š‘‘š‘„ I2=4āˆ«1ā–’1/((š‘„ āˆ’ 1)^2 āˆ’ (1)^2 āˆ’ 5) . I2=4āˆ«1ā–’1/((š‘„ āˆ’ 1)^2 āˆ’ 1 āˆ’ 5) . š‘‘š‘„ I2=4āˆ«1ā–’1/((š‘„ āˆ’ 1)^2 āˆ’ 6) . š‘‘š‘„ I2=4āˆ«1ā–’1/((š‘„ āˆ’ 1)^2 āˆ’(āˆš6 )^2 ) . š‘‘š‘„ It is of form āˆ«1ā–’š‘‘š‘„/(š‘„^2 āˆ’ š‘Ž^2 ) =1/2š‘Ž logā”|(š‘„ āˆ’ š‘Ž)/(š‘„ + š‘Ž)|+š¶ āˆ“ Replacing š‘„ by (š‘„āˆ’1) and a by āˆš6 , we get I2=4/(2āˆš6) logā”|(š‘„ āˆ’ 1 āˆ’ āˆš6)/(š‘„ āˆ’ 1 + āˆš6)|+š¶2 I2=2/āˆš6 logā”|(š‘„ āˆ’ 1 āˆ’ āˆš6)/(š‘„ āˆ’ 1 + āˆš6)|+š¶2 Putting the values of I1 and I2 in (1) āˆ«1ā–’怖(š‘„ + 2)/āˆš(š‘„^2 + 2š‘„ + 3).怗 . š‘‘š‘„ = š¼_1+š¼_2 =1/2 logā”|š‘„^2āˆ’2š‘„āˆ’5|+š¶1+2/āˆš6 logā”|(š‘„ āˆ’ 1 āˆ’ āˆš6)/(š‘„ āˆ’ 1 + āˆš6)|+š¶"2 " =šŸ/šŸ š’š’š’ˆā”|š’™^šŸāˆ’šŸš’™āˆ’šŸ“|+šŸ/āˆššŸ” š’š’š’ˆā”|(š’™ āˆ’ šŸ āˆ’ āˆššŸ”)/(š’™ āˆ’ šŸ + āˆššŸ”)|+š‘Ŗ

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