Ex 7.4, 17 - Integrate x + 2 / root x2 - 1 - Chapter 7 - Ex 7.4

Ex 7.4, 17 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.4, 17 - Chapter 7 Class 12 Integrals - Part 3
Ex 7.4, 17 - Chapter 7 Class 12 Integrals - Part 4
Ex 7.4, 17 - Chapter 7 Class 12 Integrals - Part 5


Transcript

Ex 7.4, 17 Integrate the function (π‘₯ + 2)/√(π‘₯^2 βˆ’ 1) ∫1β–’(π‘₯ + 2)/√(π‘₯^2 βˆ’ 1) . 𝑑π‘₯=∫1β–’(1/2 (2π‘₯) + 2" " )/√(π‘₯^2 βˆ’ 1) . 𝑑π‘₯ =∫1β–’(1/2 (2π‘₯))/√(π‘₯^2 βˆ’ 1) . 𝑑π‘₯+∫1β–’2/√(π‘₯^2 βˆ’ 1) . 𝑑π‘₯ =1/2 ∫1β–’( 2π‘₯)/√(π‘₯^2 βˆ’ 1) . 𝑑π‘₯+∫1β–’2/√(π‘₯^2 βˆ’ 1) . 𝑑π‘₯ Solving π‘°πŸ I1=1/2 ∫1β–’( 2π‘₯)/√(π‘₯^2 βˆ’ 1) 𝑑π‘₯ …(1) Let π‘₯^2βˆ’1=𝑑 Differentiating w.r.t. x 2π‘₯βˆ’0=𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=𝑑𝑑/2π‘₯ Thus, our equation becomes I1=1/2 ∫1β–’( 2π‘₯)/√(π‘₯^2 βˆ’ 1) 𝑑π‘₯ Put the values of (π‘₯^2βˆ’1)=𝑑 and 𝑑π‘₯, we get I1=1/2 ∫1β–’( 2π‘₯)/βˆšπ‘‘ 𝑑π‘₯ I1=1/2 ∫1β–’( 2π‘₯)/βˆšπ‘‘ Γ— 𝑑𝑑/2π‘₯ I1=1/2 ∫1β–’( 1)/βˆšπ‘‘ 𝑑𝑑 I1=1/2 ∫1β–’1/𝑑^(1/2) 𝑑𝑑 I1=1/2 ∫1▒𝑑^((βˆ’1)/2) 𝑑𝑑 I1=1/2 𝑑^((βˆ’1)/2 + 1)/((βˆ’1)/2 + 1) +𝐢1 I1=1/2 𝑑^(1/2)/(1/2) +𝐢1 I1=𝑑^(1/2)+𝐢1 I1=βˆšπ‘‘+𝐢1 I1=√(π‘₯^2 βˆ’ 1) + 𝐢1 Solving π‘°πŸ I2=∫1β–’2/√(π‘₯^2 βˆ’ 1) . 𝑑π‘₯ I2=2∫1β–’1/√(π‘₯^2 βˆ’ (1)^2 ) . 𝑑π‘₯ I2=2 π‘™π‘œπ‘”β‘|π‘₯+√(π‘₯^2 βˆ’1)|+𝐢2 It is of form ∫1▒𝑑π‘₯/√(π‘₯^2 βˆ’ π‘Ž^2 ) =π‘™π‘œπ‘”β‘|π‘₯+√(π‘₯^2 βˆ’ π‘Ž^2 )|+𝐢 ∴ Replacing a by 1 , we get ("Using " 𝑑=π‘₯^2βˆ’1) Now, Putting the values of I1 and I2 in (1) ∫1β–’(π‘₯ + 2)/√(π‘₯^2 βˆ’ 1) . 𝑑π‘₯=1/2 ∫1β–’( 2π‘₯)/√(π‘₯^2 βˆ’ 1) . 𝑑π‘₯+2∫1β–’1/√(π‘₯^2 βˆ’ 1) . 𝑑π‘₯ =√(π‘₯^2 βˆ’ 1) + 𝐢1+2 π‘™π‘œπ‘”β‘|π‘₯+√(π‘₯^2 βˆ’1) |+𝐢2 =√(𝒙^𝟐 βˆ’ 𝟏)+𝟐 π’π’π’ˆβ‘|𝒙+√(𝒙^𝟐 βˆ’πŸ)|+ π‘ͺ

Go Ad-free
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

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, Social Science, Physics, Chemistry, Computer Science at Teachoo.