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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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo