web analytics

Ex 7.4, 21 - Integrate x + 2 / root x2 + 2x + 3 - Ex 7.4

Slide8.JPG
Slide9.JPGSlide10.JPGSlide11.JPGSlide12.JPG

  1. Class 12
  2. Important Question for exams Class 12
Ask Download

Transcript

Ex 7.4, 21 (π‘₯ + 2)/√(π‘₯^2 + 2π‘₯ + 3) ∫1β–’(π‘₯ + 2)/√(π‘₯^2 + 2π‘₯ + 3) 𝑑π‘₯=1/2 ∫1β–’(2π‘₯ + 4)/√(π‘₯^2 + 2π‘₯ + 3) 𝑑π‘₯ =1/2 ∫1β–’(2π‘₯ + 2 + 4 βˆ’ 2)/√(π‘₯^2 + 2π‘₯ + 3) 𝑑π‘₯ =1/2 ∫1β–’(2π‘₯ + 2)/√(π‘₯^2 + 2π‘₯ + 3) 𝑑π‘₯+2/2 ∫1▒𝑑π‘₯/√(π‘₯^2 + 2π‘₯ + 3) 𝑑π‘₯ =1/2 ∫1β–’(2π‘₯ + 2)/√(π‘₯^2 + 2π‘₯ + 3) 𝑑π‘₯+∫1▒𝑑π‘₯/√(π‘₯^2 + 2π‘₯ + 3) 𝑑π‘₯ Solving π‘°πŸ I1=1/2 ∫1β–’(2π‘₯ + 2)/√(π‘₯^2 + 2π‘₯ + 3) . 𝑑π‘₯ Let π‘₯^2 + 2π‘₯ + 3=𝑑 Diff both sides w.r.t.x 2π‘₯+2+0=𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=𝑑𝑑/(2π‘₯ + 2) Now, our equation becomes I1=1/2 ∫1β–’(2π‘₯ + 2)/√(π‘₯^2 + 2π‘₯ + 3) . 𝑑π‘₯ Putting the value of (4π‘₯βˆ’π‘₯^2 ) and 𝑑π‘₯ I1=1/2 ∫1β–’(2π‘₯ + 2)/βˆšπ‘‘ . 𝑑π‘₯ I1=1/2 ∫1β–’(2π‘₯ + 2)/βˆšπ‘‘ . 𝑑𝑑/(2π‘₯ + 2) I1=1/2 ∫1β–’1/βˆšπ‘‘ . 𝑑𝑑 I1=1/2 ∫1β–’1/(𝑑)^(1/2) . 𝑑𝑑 I1=1/2 ∫1β–’(𝑑)^((βˆ’ 1)/2) . 𝑑𝑑 I1=1/2 〖𝑑 γ€—^((βˆ’1)/2 + 1)/((βˆ’1)/2 + 1) +𝐢1 I1= 𝑑 ^(1/2 )+𝐢1 I1= βˆšπ‘‘+𝐢1 I1=√(π‘₯^2+2π‘₯+3)+𝐢 Solving π‘°πŸ I2=∫1β–’1/√(π‘₯^2 + 2π‘₯ + 3) . 𝑑π‘₯ I2=∫1β–’1/√(π‘₯^2 + 2(π‘₯)(1) + 3) . 𝑑π‘₯ I2=∫1β–’1/√(π‘₯^2 + 2(π‘₯)(1) βˆ’ (1)^2 + (1)^2 + 3) . 𝑑π‘₯ I2=∫1β–’1/√((π‘₯ + 1)^2 βˆ’ (1)^2 + 3) . 𝑑π‘₯ I2=∫1β–’1/√((π‘₯ + 1)^2 βˆ’ 1 + 3) . 𝑑π‘₯ I2=∫1β–’1/√((π‘₯ + 1)^2 + 2) . 𝑑π‘₯ I2=∫1β–’1/√((π‘₯ + 1)^2 +(√2 )^2 ) . 𝑑π‘₯ I2=∫1β–’1/√(π‘₯^2 + 2(π‘₯)(1) βˆ’ (1)^2 + (1)^2 + 3) . 𝑑π‘₯ I2=∫1β–’1/√((π‘₯ + 1)^2 βˆ’ (1)^2 + 3) . 𝑑π‘₯ I2=∫1β–’1/√((π‘₯ + 1)^2 βˆ’ 1 + 3) . 𝑑π‘₯ I2=∫1β–’1/√((π‘₯ + 1)^2 + 2) . 𝑑π‘₯ I2=∫1β–’1/√((π‘₯ + 1)^2 +(√2 )^2 ) . 𝑑π‘₯ I2=π‘™π‘œπ‘”β‘|π‘₯+1+√((π‘₯ + 1)^2+(√2 )^2 )|+𝐢2 I2=π‘™π‘œπ‘”β‘|π‘₯+1+√(π‘₯^2+2π‘₯+1+2)|+𝐢2 I2=π‘™π‘œπ‘”β‘|π‘₯+1+√(π‘₯^2+2π‘₯+3)|+𝐢2 Putting the values of I1 and I2 in (1) ∫1β–’γ€–(π‘₯ + 2)/√(π‘₯^2 + 2π‘₯ + 3).γ€— . 𝑑π‘₯ =1/2 ∫1β–’(2π‘₯ + 2)/√(π‘₯^2 + 2π‘₯ + 3) . 𝑑π‘₯+∫1β–’1/√(π‘₯^2 + 2π‘₯ + 3) . 𝑑π‘₯ =√(π‘₯^2+2π‘₯+3)+𝐢1+π‘™π‘œπ‘”β‘|π‘₯+1+√(π‘₯^2+2π‘₯+3)|+𝐢2 =√(𝒙^𝟐+πŸπ’™+πŸ‘)+π’π’π’ˆβ‘|𝒙+𝟏+√(𝒙^𝟐+πŸπ’™+πŸ‘)|+π‘ͺ

About the Author

CA Maninder Singh's photo - Expert in Practical Accounts, Taxation and Efiling
CA Maninder Singh
CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .
Jail