Chapter 7 Class 12 Integrals

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Important Questions for exams Class 12

Ex 7.5, 11 - Chapter 7 Class 12 Integrals (Important Question)

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Ex 7.5, 11 - Integrate 5x / (x + 1) (x^2 - 4) - NCERT Maths

Ex 7.5, 11 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.5, 11 - Chapter 7 Class 12 Integrals - Part 3 Ex 7.5, 11 - Chapter 7 Class 12 Integrals - Part 4

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Ex 7.5, 11 Integrate the function 5š‘„/((š‘„ + 1) (š‘„2āˆ’ 4) ) We can write the integrand as 5š‘„/((š‘„ + 1) (š‘„2āˆ’ 4) ) = 5š‘„/((š‘„ + 1) (š‘„ āˆ’ 2) (š‘„ + 2) ) 5š‘„/((š‘„ + 1) (š‘„2āˆ’ 4) ) = š“/((š‘„ + 1) ) + šµ/((š‘„ āˆ’ 2) ) + š¶/((š‘„ + 2) ) 5š‘„/((š‘„ + 1) (š‘„2āˆ’ 4) ) = (š“(š‘„ āˆ’ 2)(š‘„ + 2) + šµ(š‘„ + 1)(š‘„ + 2) + š¶(š‘„ +1)(š‘„ āˆ’ 2))/((š‘„ + 1) (š‘„ āˆ’ 2) (š‘„ + 2) ) Cancelling denominator 5š‘„ = š“(š‘„āˆ’2)(š‘„+2)+šµ(š‘„+1)(š‘„+2)+š¶(š‘„+1)(š‘„āˆ’2) ā€¦(1) Putting x = āˆ’1 in (1) 5š‘„ = š“(š‘„āˆ’2)(š‘„+2)+šµ(š‘„+1)(š‘„+2)+š¶(š‘„+1)(š‘„āˆ’2) 5( āˆ’1) = š“(āˆ’1āˆ’2)(āˆ’1+2)+šµ(āˆ’1+1)(āˆ’1+2)+š¶(āˆ’1+1)(āˆ’1āˆ’2) āˆ’5 = š“(āˆ’3)(1)+šµĆ—0+š¶Ć—0 āˆ’5 = āˆ’3š“ š“ = (āˆ’5)/(āˆ’3) = 5/3 Putting x = 2 in (1) 5š‘„ = š“(š‘„āˆ’2)(š‘„+2)+šµ(š‘„+1)(š‘„+2)+š¶(š‘„+1)(š‘„āˆ’2) 5"(2) = " š“(2āˆ’2)(2+2)+šµ(2+1)(2+2)+š¶(2+1)(2āˆ’2) 10 = š“Ɨ0+šµ(3)(4)+š¶Ć—0 10 = 12šµ šµ = 10/12=5/6 Putting x = āˆ’2 in (1) 5š‘„ = š“(š‘„āˆ’2)(š‘„+2)+šµ(š‘„+1)(š‘„+2)+š¶(š‘„+1)(š‘„āˆ’2) 5"("āˆ’"2) = " š“(āˆ’2āˆ’2)(āˆ’2+2)+šµ(āˆ’2+1)(āˆ’2+2)+š¶(āˆ’2+1)(āˆ’2āˆ’2) āˆ’10 = š“Ɨ0+šµĆ—0+š¶(āˆ’1)(āˆ’4) āˆ’10 = 4š¶ š¶ = (āˆ’10)/4 š¶ = (āˆ’5)/2 Therefore āˆ«1ā–’5š‘„/((š‘„ + 1) (š‘„2āˆ’ 4) )=āˆ«1ā–’(š“/(š‘„ + 1)+šµ/(š‘„ āˆ’ 2)+š¶/(š‘„ + 2)) š‘‘š‘„ =5/3 āˆ«1ā–’š‘‘š‘„/(š‘„ + 1) š‘‘š‘„+ 5/6 āˆ«1ā–’š‘‘š‘„/(š‘„ āˆ’ 2) š‘‘š‘„āˆ’5/2 āˆ«1ā–’š‘‘š‘„/((š‘„ + 2) ) =šŸ“/šŸ‘ 怖š’š’š’ˆ 怗ā”|š’™+šŸ|āˆ’ šŸ“/šŸ 怖š„šØš  怗ā”|š’™+šŸ|+šŸ“/šŸ” 怖š„šØš  怗ā”|š’™āˆ’šŸ|+š‘Ŗ

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