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Chapter 7 Class 12 Integrals
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Ex 7.9, 16 - Direct Integrate 5x2 / x2 + 4x + 3 dx from 1 to 2

Ex 7.9, 16 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.9, 16 - Chapter 7 Class 12 Integrals - Part 3
Ex 7.9, 16 - Chapter 7 Class 12 Integrals - Part 4
Ex 7.9, 16 - Chapter 7 Class 12 Integrals - Part 5

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Transcript

Ex 7.9, 16 ∫_1^2β–’(5π‘₯^2)/(π‘₯^2 + 4π‘₯ + 3) 𝑑π‘₯ Let F(π‘₯)=∫1β–’γ€–(5π‘₯^2)/(π‘₯^2 + 4π‘₯ + 3) 𝑑π‘₯γ€— =5∫1β–’γ€–π‘₯^2/(π‘₯^2 + 4π‘₯ + 3) 𝑑π‘₯γ€— =5∫1β–’γ€–(π‘₯^2 + 4π‘₯ + 3 βˆ’ 4π‘₯ βˆ’ 3)/(π‘₯^2 + 4π‘₯ + 3) 𝑑π‘₯γ€— =5∫1β–’γ€–(π‘₯^2 + 4π‘₯ + 3)/(π‘₯^2 + 4π‘₯ + 3) 𝑑π‘₯γ€—βˆ’5∫1β–’γ€–( (4π‘₯ + 3))/(π‘₯^2 + 4π‘₯ + 3) 𝑑π‘₯γ€— =∫1β–’γ€–(5βˆ’(20π‘₯ + 15 )/(π‘₯^2 + 4π‘₯ + 3)) 𝑑π‘₯γ€— =∫1β–’γ€–(5βˆ’(20π‘₯ + 15 )/(π‘₯^2 + 3π‘₯ + π‘₯ + 3)) 𝑑π‘₯γ€— =∫1β–’γ€–(5βˆ’(20π‘₯ + 15 )/(π‘₯(π‘₯ + 3) + 1(π‘₯ + 3))) 𝑑π‘₯γ€— =∫1β–’γ€–(5βˆ’(20π‘₯ + 15 )/((π‘₯ + 3) (π‘₯ + 1))) 𝑑π‘₯γ€— Now, Let (20π‘₯ + 15)/(π‘₯ + 3)(π‘₯ + 1) =A/(π‘₯ + 3)+B/(π‘₯ + 1) (20π‘₯ + 15)/(π‘₯ + 3)(π‘₯ + 1) =(A(π‘₯ + 1) + B(π‘₯ + 3))/(π‘₯ + 3)(π‘₯ + 1) Canceling denominator 20π‘₯+15=A(π‘₯ + 1) + B(π‘₯ + 3) Putting π‘₯=βˆ’1 20(βˆ’1)+15=A(βˆ’1 + 1) + B(βˆ’1 + 3) βˆ’20+15=AΓ—0+B (2) Putting π‘₯=βˆ’1 20(βˆ’1)+15=A(βˆ’1 + 1) + B(βˆ’1 + 3) βˆ’20+15=AΓ—0+B (2) βˆ’5=2B B=(βˆ’5)/( 2) Similarly Putting π‘₯=βˆ’3 20(βˆ’3)+15=A(βˆ’3+1)+B(βˆ’3+3) βˆ’60+15=A(βˆ’2) BΓ—0 βˆ’45=βˆ’2A A=45/2 Hence ∫1β–’β–ˆ((5π‘₯^2)/(π‘₯^2 + 4π‘₯ + 3) " " =∫1β–’γ€–(5βˆ’(20π‘₯ + 15 )/(π‘₯^2 + 4π‘₯ + 3)) 𝑑π‘₯γ€—) =∫1β–’γ€–5βˆ’A/(π‘₯+3)βˆ’γ€— B/(π‘₯+1) 𝑑π‘₯ =∫1β–’γ€–5 𝑑π‘₯γ€—βˆ’βˆ«1β–’γ€–(45/2)/(π‘₯ +_3) 𝑑π‘₯βˆ’βˆ«1β–’γ€–(((βˆ’5)/( 2)))/(π‘₯+1) 𝑑π‘₯γ€—γ€— =5π‘₯βˆ’45/2 π‘™π‘œπ‘”|π‘₯+3|+5/2 π‘™π‘œπ‘”|π‘₯+1| Hence F(π‘₯)=5π‘₯βˆ’5/2 [9 π‘™π‘œπ‘”|π‘₯+3|βˆ’π‘™π‘œπ‘”|π‘₯+1|] Now, ∫_1^2β–’γ€–(5π‘₯^2)/(π‘₯^2 + 4π‘₯ + 3) 𝑑π‘₯=𝐹(2)βˆ’πΉ(1) γ€— =[5Γ—2βˆ’5/2 (9 π‘™π‘œπ‘”|2+3|βˆ’π‘™π‘œπ‘”|2+1|)] βˆ’ [5 Γ—1βˆ’5/2 (9 π‘™π‘œπ‘”|1+3|βˆ’π‘™π‘œπ‘”|1+1|)] =10βˆ’5/2 [9 π‘™π‘œπ‘” 5βˆ’π‘™π‘œπ‘” 3]βˆ’5+5/2 [9 π‘™π‘œπ‘” 4βˆ’π‘™π‘œπ‘” 2] =10βˆ’5βˆ’5/2 [9π‘™π‘œπ‘” 5βˆ’π‘™π‘œπ‘” 3βˆ’9π‘™π‘œπ‘” 4+π‘™π‘œπ‘” 2)] =10βˆ’5βˆ’5/2 [9π‘™π‘œπ‘” 5βˆ’9π‘™π‘œπ‘” 4βˆ’(π‘™π‘œπ‘” 3βˆ’π‘™π‘œπ‘” 2)] =πŸ“βˆ’πŸ“/𝟐 (πŸ— π₯𝐨𝐠 πŸ“/πŸ’βˆ’π₯𝐨𝐠 πŸ‘/𝟐) (log a βˆ’ log b = log π‘Ž/𝑏)

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.